Past JEE Main Entrance Paper

Hint

(a) \(3(\sin \theta-\cos \theta)^{4}+6(\sin \theta+\cos \theta)^{2}+4 \sin ^{6} \theta\)
\(
\begin{aligned}
&=3(1-2 \sin \theta \cos \theta)^{2}+6(1+2 \sin \theta \cos \theta)+4 \sin ^{6} \theta \\
&=3\left(1+4 \sin ^{2} \theta \cos ^{2} \theta-4 \sin \theta \cos \theta\right)+6 -12 \sin \theta \cos \theta+4 \sin ^{6} \theta\\
&=9+12 \sin ^{2} \theta \cos ^{2} \theta+4 \sin ^{6} \theta \\
&=9+12 \cos ^{2} \theta\left(1-\cos ^{2} \theta\right)+4\left(1-\cos ^{2} \theta\right)^{3} \\
&=9+12 \cos ^{2} \theta-12 \cos ^{4} \theta+4\left(1-\cos ^{6} \theta-3 \cos ^{2} \theta+3 \cos ^{4} \theta\right) \\
&=9+4-4 \cos ^{6} \theta \\
&=13-4 \cos ^{6} \theta
\end{aligned}
\)

Hint

(b) Let \(f_{k}(x)=\frac{1}{k}\left(\sin ^{k} x+\cos ^{k} x\right)\)
Consider \(f_{4}(x)-f_{6}(x)=\frac{1}{4}\left(\sin ^{4} x+\cos ^{4} x\right)\)
\(
\begin{aligned}
&-\frac{1}{6}\left(\sin ^{6} x+\cos ^{6} x\right) \\
&=\frac{1}{4}\left[1-2 \sin ^{2} x \cos ^{2} x\right]-\frac{1}{6}\left[1-3 \sin ^{2} x \cos ^{2} x\right] \\
&=\frac{1}{4}-\frac{1}{6}=\frac{1}{12}
\end{aligned}
\)

Hint

(d) Given \(2 \cos \theta+\sin \theta=1\)
Squaring both sides, we get
\(
\begin{aligned}
&(2 \cos \theta+\sin \theta)^{2}=1^{2} \\
&\Rightarrow 4 \cos ^{2} \theta+\sin ^{2} \theta+4 \sin \theta \cos \theta=1 \\
&\Rightarrow 3 \cos ^{2} \theta+\left(\cos ^{2} \theta+\sin ^{2} \theta\right)+4 \sin \theta \cos \theta=1
\end{aligned}
\)
\(\begin{aligned}
&\Rightarrow 3 \cos ^{2} \theta+4 \sin \theta \cos \theta=0\\
&\Rightarrow 3 \cos ^{2} \theta+4 \sin \theta \cos \theta=0\\
&\Rightarrow \cos \theta(3 \cos \theta+4 \sin \theta)=0\\
&\Rightarrow 3 \cos \theta+4 \sin \theta=0 \Rightarrow 3 \cos \theta=-4 \sin \theta\\
&\Rightarrow \frac{-3}{4}=\tan \theta=\sqrt{\sec ^{2} \theta-1}=\frac{-3}{4}\left(\because \tan \theta=\sqrt{\sec ^{2} \theta-1}\right)\\
&\Rightarrow \sec ^{2} \theta-1=\left(\frac{-3}{4}\right)^{2}=\frac{9}{16}\\
&\Rightarrow \sec ^{2} \theta=\frac{9}{16}+1=\frac{25}{16} \Rightarrow \sec \theta=\frac{5}{4}\\
&\text { or } \cos \theta=\frac{4}{5}\\
&\sin ^{2} \theta+\frac{16}{25}=1 \Rightarrow \sin ^{2} \theta=1-\frac{16}{25}=\frac{9}{25}\\
&\sin \theta=\pm \frac{3}{5}\\
&=7 \times \frac{4}{5}+6 \times \frac{3}{5}=\frac{28}{5}+\frac{18}{5}=\frac{46}{5}
\end{aligned}\)

Hint

(b) Given expression can be written as
\(
\begin{aligned}
\frac{\sin A}{\cos A} \times \frac{\sin A}{\sin A-\cos A}+\frac{\cos A}{\sin A} \times \frac{\cos A}{\cos A-\sin A} \\
&\left(\because \tan A=\frac{\sin A}{\cos A} \text { and } \cot A=\frac{\cos A}{\sin A}\right)
\end{aligned}
\)
\(\begin{aligned}
&=\frac{1}{\sin A-\cos A}\left\{\frac{\sin ^{3} A-\cos ^{3} A}{\cos A \sin A}\right\} \\
&\because a^{3}-b^{3}=(a-b)\left(a^{2}+a b+b^{2}\right) \\
&=\frac{\sin ^{2} A+\sin A \cos A+\cos ^{2} A}{\sin A \cos A} \\
&=1+\sec \mathrm{A} \operatorname{cosec} \mathrm{A}
\end{aligned}\)

Hint

(b) Given : \(\sin \theta=1 / 2\) and \(\theta\) is acute angle
\(
\therefore \theta=\pi / 6
\)
Also given, \(\cos \phi=\frac{1}{2}\) and \(\varphi\) is acute angle.
\(
\begin{aligned}
&\therefore 0<\frac{1}{3}<\frac{1}{2} \\
&\Rightarrow \cos \pi / 2<\cos \phi<\cos \pi / 3 \text { or } \pi / 3<\phi<\pi / 2 \\
&\therefore \frac{\pi}{3}+\frac{\pi}{6}<\theta+\phi<\frac{\pi}{2}+\frac{\pi}{6} \text { or } \frac{\pi}{2}<\theta+\phi<\frac{2 \pi}{3} \\
&\Rightarrow \theta+\phi \in\left(\frac{\pi}{2}, \frac{2 \pi}{3}\right)
\end{aligned}
\)

Hint

\(\begin{aligned}
&\text { (c) } \sin \left\{\left(\omega^{10}+\omega^{23}\right) \pi-\frac{\pi}{4}\right\}=\sin \left\{\left(\omega+\omega^{2}\right) \pi-\frac{\pi}{4}\right\} \\
&=\sin \left(-\pi-\frac{\pi}{4}\right)=-\sin \left(\pi+\frac{\pi}{4}\right)=\sin \pi / 4=1 / \sqrt{2}
\end{aligned}\)

Hint

(b) \(\tan \theta=\frac{-4}{3} \Rightarrow \theta \in\) II quad or IV quad.
\(
\begin{aligned}
&\Rightarrow 0<\sin \theta<1 \text { or }-1<\sin \theta<0 \\
&\Rightarrow \sin \theta=\frac{4}{5} \text { or }-\frac{4}{5}
\end{aligned}
\)

Hint

(c) We know, \(\sin 15^{\circ}=\frac{\sqrt{3}-1}{2 \sqrt{2}}\) (irrational) \(\cos 15^{\circ}=\frac{\sqrt{3}+1}{2 \sqrt{2}}(\) irrational \()\)
\(\begin{aligned}
&\sin 15^{\circ} \cdot \cos 15^{\circ}=\frac{1}{2}\left(2 \sin 15^{\circ} \cos 15^{\circ}\right) \\
&=\frac{1}{2} \sin 30^{\circ}=\frac{1}{4}(\text { rational }) \\
&\sin 15^{\circ} \cos 75^{\circ}=\sin 15^{\circ} \cos \left(90-15^{\circ}\right) \\
&=\sin 15^{\circ} \sin 15^{\circ}=\sin ^{2} 15^{\circ}=\frac{1}{2}\left(1+\cos 30^{\circ}\right) \\
&=\frac{1}{2}\left(1-\frac{\sqrt{3}}{2}\right) \text { (irrational) }
\end{aligned}\)

Hint

(a) \((\mathrm{A} \rightarrow \mathrm{r} ; \mathrm{B} \rightarrow \mathbf{p})\)

(p) If \(\frac{13 \pi}{48}<\alpha<\frac{14 \pi}{48}\) then \(\frac{13 \pi}{16}<3 \alpha<\frac{14 \pi}{16}\)
and \(\frac{13 \pi}{24}<2 \alpha<\frac{14 \pi}{24}\)
\(\Rightarrow 3 \alpha \in\) II quad and \(2 \alpha \in\) II quad \(\Rightarrow \sin 3 \alpha=+\) ve
\(\cos 2 \alpha=-\mathrm{ve} \therefore \frac{\sin 3 \alpha}{\cos 2 \alpha}=-v e\)
\(\therefore\) (B) corresponds to \((\mathrm{p})\).
(q) If \(\alpha \in\left(\frac{14 \pi}{48}, \frac{18 \pi}{48}\right)\) then \(\frac{14 \pi}{16}<3 \alpha<\frac{18 \pi}{16}\)
and \(\frac{14 \pi}{24}<2 \alpha<\frac{18 \pi}{24}\)
\(\Rightarrow 3 \alpha \in\) II or III quad and \(2 \alpha \in\) II quad
\(\Rightarrow\) Nothing can be said about the sign of \(\frac{\sin 3 \alpha}{\cos 2 \alpha}\) over this interval.
(r) If \(\alpha \in\left(\frac{18 \pi}{48}, \frac{23 \pi}{48}\right)\) then \(\frac{18 \pi}{16}<3 \alpha<\frac{23 \pi}{16}\)
and \(\frac{18 \pi}{24}<2 \alpha<\frac{23 \pi}{24}\)
\(\Rightarrow 3 \alpha \in\) III quad and \(2 \alpha \in\) II quad
\(\Rightarrow \quad \sin 3 \alpha=-\mathrm{ve}, \cos 2 \mathrm{a}=-\mathrm{ve}, \quad \therefore \frac{\sin 3 \alpha}{\cos 2 \alpha}=+v e\)
\(\therefore\) (A) corresponds to (r)
(s) If \(\alpha \in(0, \pi / 2)\), then \(0<3 \alpha<3 \pi / 2\) and \(0<2 \alpha<\pi\)
\(\Rightarrow\) Nothing can be said about the sign of \(\frac{\sin 3 \alpha}{\cos 2 \alpha}\) over the given interval.

Hint

(a) \(\overrightarrow{\mathrm{OX}}, \overrightarrow{\mathrm{OY}}, \overrightarrow{\mathrm{OZ}}\) are unit vectors in the directions of sides \(\overrightarrow{\mathrm{OR}}, \overrightarrow{\mathrm{RP}}\) and \(\overrightarrow{\mathrm{PQ}}\) respectively,

\(\begin{aligned}
&\therefore \quad \overrightarrow{\mathrm{OX}}=\frac{\overrightarrow{\mathrm{QR}}}{|\overrightarrow{\mathrm{QR}}|}, \overrightarrow{\mathrm{OY}}=\frac{\overrightarrow{\mathrm{RP}}}{|\overrightarrow{\mathrm{RP}}|}, \overrightarrow{\mathrm{OZ}}=\frac{\overrightarrow{\mathrm{PQ}}}{|\overrightarrow{\mathrm{PQ}}|} \\
&\therefore \quad|\overrightarrow{\mathrm{OX}} \times \overrightarrow{\mathrm{OY}}|=\frac{|\overrightarrow{\mathrm{OR}} \times \overrightarrow{\mathrm{RP}}|}{|\overrightarrow{\mathrm{QR}}||\overrightarrow{\mathrm{RP}}|}=\frac{|\overrightarrow{\mathrm{QR}}||\overrightarrow{\mathrm{RP}}| \sin (\mathrm{P}+\mathrm{Q})}{|\overrightarrow{\mathrm{QR}}||\overrightarrow{\mathrm{RP}}|} \\
&=\sin (\mathrm{P}+\mathrm{Q})
\end{aligned}\)

Hint

(b) \(\cos (\mathrm{P}+\mathrm{Q})+\cos (\mathrm{Q}+\mathrm{R})+\cos (\mathrm{R}+\mathrm{P})\)
\(=\cos (180-\mathrm{R})+\cos (180-\mathrm{P})+\cos (180-\mathrm{Q})\)
\(=-[\cos P+\cos Q+\cos R]\)
In any \(\mathrm{PQR}, \cos P+\cos Q+\cos R \leq \frac{3}{2}\)
\(\Rightarrow \quad-(\cos P+\cos Q+\cos R) \geq-\frac{3}{2}\)
\(\therefore \quad\) Required minimum value \(=-\frac{3}{2}\)

Hint

(a) \(\text { Given : } 2 \sin t=\frac{1-2 x+5 x^{2}}{3 x^{2}-2 x-1}, \quad t \in[-\pi / 2, \pi / 2]\)
\(\Rightarrow(6 \sin t-5) x^{2}+2(1-2 \sin t) x-(1+2 \sin t)=0\)
The given equation will hold, if \(x\) be some real number, and hence, \(D \geq 0\)
\(
\begin{aligned}
&\Rightarrow 4(1-2 \sin t)^{2}+4(6 \sin t-5)(1+2 \sin t) \geq 0 \\
&\Rightarrow 16 \sin ^{2} t-8 \sin t-4 \geq 0 \Rightarrow\left(4 \sin ^{2} t-2 \sin t-1\right) \geq 0 \\
&\Rightarrow 4\left(\sin t-\frac{\sqrt{5}+1}{4}\right)\left(\sin t+\frac{\sqrt{5}-1}{4}\right) \geq 0 \\
&\Rightarrow \sin t \leq-\left(\frac{\sqrt{5}-1}{4}\right) \text { or } \sin t \geq \frac{\sqrt{5}+1}{4} \\
&\Rightarrow \sin t \leq \sin (-\pi / 10) \text { or } \quad \sin t \geq \sin (3 \pi / 10) \\
&\Rightarrow t \leq-\pi / 10 \quad \text { or } \quad t \geq 3 \pi / 10 \\
&\text { [Note that } \sin x \text { is an increasing function from }-\pi / 2 \text { to } \pi / 2] \\
&\therefore \quad \text { range of } t \text { is }[-\pi / 2,-\pi / 10] \cup[3 \pi / 10, \pi / 2]
\end{aligned}
\)

Hint

(d) \(\mathrm{L}+\mathrm{M}=1-2 \sin ^{2} \frac{\pi}{8}=\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}\dots….(i)\)
and \(\mathrm{L}-\mathrm{M}=-\cos \frac{\pi}{8} \dots….(ii)\)
From equations (i) and (ii),
\(
\begin{aligned}
&L=\frac{1}{2}\left(\frac{1}{\sqrt{2}}-\cos \frac{\pi}{8}\right)=\frac{1}{2 \sqrt{2}}-\frac{1}{2} \cos \frac{\pi}{8} \text { and } \\
&M=\frac{1}{2}\left(\frac{1}{\sqrt{2}}+\cos \frac{\pi}{8}\right)=\frac{1}{2 \sqrt{2}}+\frac{1}{2} \cos \frac{\pi}{8}
\end{aligned}
\)

Hint

(a) Let \(f(x, y)=x+y-1\)
Given \((1,2)\) and \((\sin \theta, \cos \theta)\) are lies on same side.
\(
\begin{aligned}
&\therefore f(1,2) \cdot f(\sin \theta, \cos \theta)>0 \\
&\Rightarrow 2[\sin \theta+\cos \theta-1]>0 \\
&\Rightarrow \sin \theta+\cos \theta>1 \Rightarrow \sin \left(\theta+\frac{\pi}{4}\right)>\frac{1}{\sqrt{2}} \\
&\Rightarrow \theta+\frac{\pi}{4} \in\left(\frac{\pi}{4}, \frac{3 \pi}{4}\right) \Rightarrow \theta \in\left(0, \frac{\pi}{2}\right)
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (b) } \cos ^{3} \frac{\pi}{8}\left[4 \cos ^{3} \frac{\pi}{8}-3 \cos \frac{\pi}{8}\right]\\
&+\sin ^{3} \frac{\pi}{8}\left[3 \sin \frac{\pi}{8}-4 \sin ^{3} \frac{\pi}{8}\right]\\
&=4 \cos ^{6} \frac{\pi}{8}-4 \sin ^{6} \frac{\pi}{8}-3 \cos ^{4} \frac{\pi}{8}+3 \sin ^{4} \frac{\pi}{8}\\
&=4\left[\left(\cos ^{2} \frac{\pi}{8}-\sin ^{2} \frac{\pi}{8}\right)\right]\\
&\left[\left(\sin ^{4} \frac{\pi}{8}+\cos ^{4} \frac{\pi}{8}+\sin ^{2} \frac{\pi}{8} \cos ^{2} \frac{\pi}{8}\right)\right]\\
&-3\left[\left(\cos ^{2} \frac{\pi}{8}-\sin ^{2} \frac{\pi}{8}\right)\left(\cos ^{2} \frac{\pi}{8}+\sin ^{2} \frac{\pi}{8}\right)\right]\\
&=\cos \frac{\pi}{4}\left[4\left(1-\sin ^{2} \frac{\pi}{8} \cos ^{2} \frac{\pi}{8}\right)-3\right]\\
&=\frac{1}{\sqrt{2}}\left[1-\frac{1}{2}\right]=\frac{1}{2 \sqrt{2}}
\end{aligned}
\)

Hint

(a) Let the height of the tower be \(h\) and the distance of the foot of the tower from the point \(A\) is \(d\).
By the diagram,

\(
\begin{gathered}
\tan 45^{\circ}=\frac{h}{d}=1 \\
h=d \dots ….(i)
\end{gathered}
\)
\(\tan 30^{\circ}=\frac{h-30}{d}\)
\(
\sqrt{3}(h-30)=d \dots ….(ii)
\)
Put the value of \(h\) from (i) to (ii),
\(
\begin{aligned}
&\sqrt{3} d=d+30 \sqrt{3} \\
&d=\frac{30 \sqrt{3}}{\sqrt{3-1}}=15 \sqrt{3}(\sqrt{3}+1)=15(3+\sqrt{3})
\end{aligned}
\)

Hint

(b) \(\cos ^{2} 10^{\circ}-\cos 10^{\circ} \cos 50^{\circ}+\cos ^{2} 50^{\circ}\)
\(
\begin{aligned}
&=\left(\frac{1+\cos 20^{\circ}}{2}\right)+\left(\frac{1+\cos 100^{\circ}}{2}\right)-\frac{1}{2}\left(2 \cos 10^{\circ} \cos 50^{\circ}\right) \\
&=1+\frac{1}{2}\left(\cos 20^{\circ}+\cos 100^{\circ}\right)-\frac{1}{2}\left[\cos 60^{\circ}+\cos 40^{\circ}\right] \\
&=\left(1-\frac{1}{4}\right)+\frac{1}{2}\left[\cos 20^{\circ}+\cos 100^{\circ}-\cos 40^{\circ}\right] \\
&=\frac{3}{4}+\frac{1}{2}\left[2 \cos 60^{\circ} \times \cos 40^{\circ}-\cos 40^{\circ}\right] \\
&=\frac{3}{4}
\end{aligned}
\)

Hint

(b) \(\alpha+\beta\) and \(\alpha-\beta\) both are acute angles.
\(\cos (\alpha+\beta)=\frac{3}{5}\), then \(\sin (\alpha+\beta)=\sqrt{1-\left(\frac{3}{5}\right)^{2}}=\frac{4}{5}\)
\(\tan (\alpha+\beta)=\frac{4}{3}\)
And \(\sin (\alpha-\beta)=\frac{5}{13}\), then
\(\cos (\alpha-\beta)=\sqrt{1-\left(\frac{5}{13}\right)^{2}}=\frac{12}{13}\)
\(\Rightarrow \tan (\alpha-\beta)=\frac{5}{12}\)
Now, \(\tan 2 \alpha=\tan ((\alpha+\beta)+(\alpha-\beta))\)
\(=\frac{\tan (\alpha+\beta)+\tan (\alpha-\beta)}{1-\tan (\alpha+\beta) \cdot \tan (\alpha-\beta)}=\frac{\frac{4}{3}+\frac{5}{12}}{1-\frac{4}{3} \cdot \frac{5}{12}}=\frac{63}{16}\)

Hint

(d)

\(
\text { A.M. } \geq \text { G.M. }
\)
\(\frac{\sin ^{4} \alpha+4 \cos ^{4} \beta+1+1}{4} \geq\left(\sin ^{4} \alpha \cdot 4 \cdot \cos ^{4} \beta .1 .1\right) \frac{1}{4}\)
\(
\begin{aligned}
&\Rightarrow \text { A.M. }=\text { G.M. } \Rightarrow \sin ^{4} \alpha=1=4 \cos ^{4} \beta \\
&\sin \alpha=1, \cos \beta=\pm \frac{1}{\sqrt{2}} \\
&\sin \beta=\frac{1}{\sqrt{2}} \operatorname{as} \beta \in[0, \pi] \\
&\cos (\alpha+\beta)-\cos (\alpha-\beta)=-2 \sin \alpha \sin \beta \\
&=-\sqrt{2}
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (a) } f_{k}(x)=\frac{1}{k}\left(\sin ^{k} x+\cos ^{k} x\right)\\
&f_{4}(x)=\frac{1}{4}\left[\sin ^{4} x+\cos ^{4} x\right]\\
&=\frac{1}{4}\left[\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-\frac{(\sin 2 x)^{2}}{2}\right]\\
&=\frac{1}{4}\left[1-\frac{(\sin 2 x)^{2}}{2}\right]
\end{aligned}
\)
\(
\begin{aligned}
& f_{6}(x)=\frac{1}{6}\left[\sin ^{6} x+\cos ^{6} x\right] \\
=& \frac{1}{6}\left[\left(\sin ^{2} x+\left(\cos ^{2} x\right)-\frac{3}{4}\left(\sin ^{2} x\right)^{2}\right]\right.\\
=& \frac{1}{6}\left[1-\frac{3}{4}(\sin 2 x)^{2}\right] \\
& \text { Now } f_{4}(x)-f_{(6)}(x)=\frac{1}{4}-\frac{1}{6}-\frac{(\sin 2 x)^{2}}{8}+\frac{1}{8}(\sin 2 x)^{2} \\
=& \frac{1}{12}
\end{aligned}
\)

Hint

(a) We have
\(
\begin{aligned}
&5 \tan ^{2} \mathrm{x}-5 \cos ^{2} \mathrm{x}=2\left(2 \cos ^{2} \mathrm{x}-1\right)+9 \\
&\quad \Rightarrow 5 \tan ^{2} \mathrm{x}-5 \cos ^{2} \mathrm{x}=4 \cos ^{2} \mathrm{x}-2+9 \\
&\quad \Rightarrow 5 \tan ^{2} \mathrm{x}=9 \cos ^{2} \mathrm{x}+7 \\
&\quad \Rightarrow 5\left(\sec ^{2} \mathrm{x}-1\right)=9 \cos ^{2} \mathrm{x}+7 \\
&\text { Let } \cos ^{2} \mathrm{x}=\mathrm{t} \\
&\quad \Rightarrow \frac{5}{\mathrm{t}}-9 \mathrm{t}-12=0 \\
&\Rightarrow 9 \mathrm{t}^{2}+12 \mathrm{t}-5=0 \\
&\quad \Rightarrow 9 \mathrm{t}^{2}+15 \mathrm{t}-3 \mathrm{t}-5=0 \\
&\Rightarrow(3 \mathrm{t}-1)(3 \mathrm{t}+5)=0 \\
&\Rightarrow \mathrm{t}=\frac{1}{3} \text { as } \mathrm{t} \neq-\frac{5}{3} \\
&\cos 2 \mathrm{x}=2 \cos ^{2} \mathrm{x}-1=2\left(\frac{1}{3}\right)-1=-\frac{1}{3} \\
&\cos 4 \mathrm{x}=2 \cos ^{2} 2 \mathrm{x}-1=2\left(-\frac{1}{3}\right)^{2}-1=-\frac{7}{9}
\end{aligned}
\)

Hint

(b) \(4+\frac{1}{2} \sin ^{2} 2 x-2 \cos ^{4} x\)
\(
=4+2\left(1-\cos ^{2} x\right) \cos ^{2} x-2 \cos ^{4} x
\)
\(
\begin{aligned}
=-4 &\left\{\cos ^{4} x-\frac{\cos ^{2} x}{2}-1+\frac{1}{16}-\frac{1}{16}\right\} \\
=-4 &\left\{\left(\cos ^{2} x-\frac{1}{4}\right)^{2}-\frac{17}{16}\right\} \\
0 \leq \cos ^{2} x & \leq 1 \\
-\frac{1}{4} & \leq \cos ^{2} x-\frac{1}{4} \leq \frac{3}{4} \\
0 & \leq\left(\cos ^{2} x-\frac{1}{4}\right)^{2} \leq \frac{9}{16} \\
-\frac{17}{16} & \leq\left(\cos ^{2} x-\frac{1}{4}\right)^{2}-\frac{17}{16} \leq \frac{9}{16}-\frac{17}{16} \\
\frac{17}{4} & \geq-4\left\{\left(\cos ^{2} x-\frac{1}{4}\right)^{2}-\frac{17}{16}\right\} \geq \frac{1}{2} \\
M &=\frac{17}{4} \\
m &=\frac{1}{2} \\
M-m &=\frac{17}{4}-\frac{2}{4}=\frac{15}{4}
\end{aligned}
\)

Hint

(c) 

\(
\begin{aligned}
& 2 \sum_{k=1}^{13} \frac{\sin \left(\frac{\pi}{6}\right)}{\sin \left(\frac{\pi}{4}+\frac{(k-1) \pi}{6}\right) \sin \left(\frac{\pi}{4}+\frac{k \pi}{6}\right)} \\
& =2 \sum \frac{\sin \left\{\left(\frac{\pi}{4}+\frac{k \pi}{6}\right)-\left(\frac{\pi}{4}+\frac{(k-1) \pi}{6}\right)\right\}}{\sin \left(\frac{\pi}{4}+\frac{(k-1) \pi}{6}\right) \cdots \in\left(\frac{\pi}{4}+\frac{k \pi}{6}\right)} \\
& =2 \sum_{k=1}^{13}\left\{\cot \left(\frac{\pi}{4}+\frac{(k-1) \pi}{6}\right)-\cot \left(\frac{\pi}{4}+\frac{k \pi}{6}\right)\right\} \\
& =2\left[\cot \left(\frac{\pi}{4}\right)-\cot \left(\frac{\pi}{4}+\frac{13 \pi}{6}\right)\right] \\
& =2[1-(2-\sqrt{3})] \\
& =2(\sqrt{3}-1)
\end{aligned}
\)

Hint

(b) Let \(\cos \alpha+\cos \beta=\frac{3}{2}\)
\(\Rightarrow 2 \cos \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}=\frac{3}{2}\)
and \(\sin \alpha+\sin \beta=\frac{1}{2}\)
\(
\Rightarrow 2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}=\frac{1}{2}
\)
On dividing (ii) by (i), we get
\(
\tan \left(\frac{\alpha+\beta}{2}\right)=\frac{1}{3}
\)
Given : \(\theta=\frac{\alpha+\beta}{2} \Rightarrow 2 \theta=\alpha+\beta\)
Consider \(\sin 2 \theta+\cos 2 \theta=\sin (\alpha+\beta)+\cos (\alpha+\beta)\)
\(
=\frac{\frac{2}{3}}{1+\frac{1}{9}}+\frac{1-\frac{1}{9}}{1+\frac{1}{9}}=\frac{6}{10}+\frac{8}{10}=\frac{7}{5}
\)

Hint

(b) Given: \(\theta \in\left(0, \frac{\pi}{4}\right) \Rightarrow \tan \theta<1\) and \(\cot \theta>1\)
Let \(\tan \theta=1-x\) and \(\cot \theta=1+y\),
where \(x, y>0\) and are very small, then
\(
\begin{aligned}
&\therefore t_{1}=(1-x)^{1-x}, t_{2}=(1-x)^{1+y} \\
&\mathrm{t}_{3}=(1+y)^{1-x}, t_{4}=(1+y)^{1+y} \\
&\text { Clearly, } t_{4}>t_{3} \text { and } t_{1}>t_{2} \text { also, } t_{3}>t_{1} \\
&\therefore t_{4}>t_{3}>t_{1}>t_{2}
\end{aligned}
\)

Hint

(a) \(2 \sin ^{2} \theta-5 \sin \theta+2>0\)
\(
\begin{aligned}
&\Rightarrow(\sin \theta-2)(2 \sin \theta-1)>0 \\
&\because-1 \leq \sin \theta \leq 1, \quad \therefore \sin \theta<\frac{1}{2}
\end{aligned}
\)

\(
\text { From graph, we get } x \in\left(0, \frac{\pi}{6}\right) \cup\left(\frac{5 \pi}{6}, 2 \pi\right)
\)

Hint

(c) Given : \(\alpha+\beta=\pi / 2 \Rightarrow \alpha=\pi / 2-\beta\)
\(
\begin{aligned}
&\Rightarrow \tan \alpha=\tan (\pi / 2-\beta)=\cot \beta=\frac{1}{\tan \beta} \\
&\Rightarrow \tan \alpha \tan \beta=1 \Rightarrow 1+\tan \alpha \tan \beta=2 .
\end{aligned}
\)
Now, \(\tan (\alpha-\beta)=\frac{\tan \alpha-\tan \beta}{1+\tan \alpha \tan \beta}\)
\(
\begin{aligned}
&\Rightarrow \tan \gamma=\frac{\tan \alpha-\tan \beta}{2} \\
&\Rightarrow 2 \tan \gamma=\tan \alpha-\tan \beta \Rightarrow \tan \alpha=2 \tan \gamma+\tan \beta
\end{aligned}
\)

Hint

(a) Given : \(\left(\cot \alpha_{1}\right) \cdot\left(\cot \alpha_{2}\right) \ldots\left(\cot \alpha_{n}\right)=1\)
\(
\begin{aligned}
\Rightarrow\left(\cos \alpha_{1}\right)\left(\cos \alpha_{2}\right) \ldots\left(\cos \alpha_{n}\right) \\
=&\left(\sin \alpha_{1}\right)\left(\sin \alpha_{2}\right) \ldots\left(\sin \alpha_{n}\right) \dots ….(i)
\end{aligned}
\)
Let \(\mathrm{y}=\left(\cos \alpha_{1}\right)\left(\cos \alpha_{2}\right) \ldots\left(\cos \alpha_{n}\right)\) (to be max.)
\(
\Rightarrow \mathrm{y}^{2}=\left(\cos ^{2} \alpha_{1}\right)\left(\cos ^{2} \alpha_{2}\right) \ldots\left(\cos ^{2} \alpha_{n}\right)
\)
\(
\begin{aligned}
&=\cos \alpha_{1} \sin \alpha_{1} \cos \alpha_{2} \sin \alpha_{2} \ldots \cos \alpha_{n} \sin \alpha_{n} \quad(\text { From(i)) } \\
&=\frac{1}{2^{n}}\left[\sin 2 \alpha_{1} \sin 2 \alpha_{2} \ldots . \sin 2 \alpha_{n}\right] \\
&\text { Now, } 0 \leq \alpha_{1}, \alpha_{2}, \ldots \ldots \alpha_{n} \leq \pi / 2 \\
&\therefore 0 \leq 2 \alpha_{1}, 2 \alpha_{2}, \ldots \ldots, 2 \alpha_{n} \leq \pi \\
&\Rightarrow 0 \leq \sin 2 \alpha_{1}, \sin 2 \alpha_{2}, \ldots ., \sin 2 \alpha_{n} \leq 1 \\
&\therefore y^{2} \leq \frac{1}{2^{n}} \cdot 1 \Rightarrow y \leq \frac{1}{2^{n / 2}} \\
&\therefore \text { Max. value of } y \text { i.e. }\left(\cos \alpha_{1}\right) \cdot\left(\cos \alpha_{2}\right) \ldots .\left(\cos \alpha_{n}\right)=\frac{1}{2^{n / 2}} .
\end{aligned}
\)

Hint

(c) \(f(\theta)=\sin \theta(\sin \theta+\sin 3 \theta)\)
\(
\begin{aligned}
&=\sin \theta\left(\sin \theta+3 \sin \theta-4 \sin ^{3} \theta\right) \\
&=\sin \theta\left(4 \sin \theta-4 \sin ^{3} \theta\right)=\sin ^{2} \theta\left(4-4 \sin ^{2} \theta\right) \\
&=4 \sin ^{2} \theta\left(1-\sin ^{2} \theta\right) \\
&=4 \sin ^{2} \theta \cos ^{2} \theta=(2 \sin \theta \cos \theta)^{2}=(\sin 2 \theta)^{2} \geq 0,
\end{aligned}
\)
which is true for all \(\theta\).

Hint

\(
\begin{aligned}
&\text { (c) } 3(\sin x-\cos x)^{4}+6(\sin x+\cos x)^{2}+4\left(\sin ^{6} x+\cos ^{6} x\right) \\
&=3(1-\sin 2 x)^{2}+6(1+\sin 2 x) \\
&\qquad+4\left[\left(\sin ^{2} x+\cos ^{2} x\right)^{3}-3 \sin ^{2} x \cos ^{2} x\left(\sin ^{2} x+\cos ^{2} x\right)\right]
\end{aligned}
\)

Hint

(b) \(\sec 2 \mathrm{x}-\tan 2 \mathrm{x}=\frac{1-\sin 2 x}{\cos 2 x}=\frac{1-\cos 2\left(\frac{\pi}{4}-x\right)}{\sin 2\left(\frac{\pi}{4}-x\right)}\)
\(
=\frac{2 \sin ^{2}\left(\frac{\pi}{4}-x\right)}{2 \sin \left(\frac{\pi}{4}-x\right) \cos \left(\frac{\pi}{4}-x\right)}=\tan \left(\frac{\pi}{4}-x\right)
\)

Hint

\(
\text { (b) } A=\sin ^{2} \theta+\cos ^{4} \theta=\sin ^{2} \theta+\left(1-\sin ^{2} \theta\right)^{2}
\)
\(
=\sin ^{4} \theta-\sin ^{2} \theta+1 \Rightarrow A=\left(\sin ^{2} \theta-\frac{1}{2}\right)^{2}+\frac{3}{4}
\)
Now,
\(
\begin{aligned}
&0 \leq\left(\sin ^{2} \theta-\frac{1}{2}\right)^{2} \leq \frac{1}{4} \\
&\Rightarrow \frac{3}{4} \leq\left(\sin ^{2} \theta-\frac{1}{2}\right)^{2}+\frac{3}{4} \leq 1 \Rightarrow \frac{3}{4} \leq A \leq 1
\end{aligned}
\)

Hint

(a) \(\alpha+\beta+\gamma=2 \pi \Rightarrow \frac{\alpha}{2}+\frac{\beta}{2}+\frac{\gamma}{2}=\pi\)
\(
\begin{gathered}
\therefore \tan \left(\frac{\alpha}{2}+\frac{\beta}{2}\right)=\tan \left(\pi-\frac{\gamma}{2}\right)=-\tan \frac{\gamma}{2} \\
\Rightarrow \frac{\tan \alpha / 2+\tan \beta / 2}{1-\tan \alpha / 2 \tan \beta / 2}=-\tan \gamma / 2 \\
\Rightarrow \tan \alpha / 2+\tan \beta / 2+\tan \gamma / 2 \\
=\tan \alpha / 2 \tan \beta / 2 \tan \gamma / 2
\end{gathered}
\)

Hint

(b) Let \(f(\theta)=\frac{1}{g(\theta)}\),
where \(g(\theta)=\sin ^{2} \theta+3 \sin \theta \cos \theta+5 \cos ^{2} \theta\)
Clearly \(f\) is maximum when \(g\) is minimum
Now \(g(\theta)=\frac{1-\cos 2 \theta}{2}+\frac{3}{2} \sin 2 \theta+\frac{5}{2}(1+\cos 2 \theta)\)
\(
\begin{aligned}
&=3+2 \cos 2 \theta+\frac{3}{2} \sin 2 \theta \geq 3+\left(-\sqrt{4+\frac{9}{4}}\right) \\
&\therefore g_{\min }=3-\frac{5}{2}=\frac{1}{2} \Rightarrow f_{\max }=2 .
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (c) } \frac{\sqrt{2} \sin \alpha}{\sqrt{2} \cos \alpha} = \frac{1}{\sqrt{7}} \text { and } \sqrt{\frac{1-\cos {2} \beta}{2}}=\frac{1}{10} \\
&\Rightarrow \frac{\sqrt{2} \sin \beta}{\sqrt{2}}=\frac{1}{\sqrt{10}}
\end{aligned}
\)
\(
\begin{gathered}
\therefore \quad \tan \alpha=\frac{1}{7} \text { and } \sin \beta=\frac{1}{\sqrt{10}} \\
\tan \beta=\frac{1}{3} \\
\therefore \quad \tan 2 \beta=\frac{2 \tan \beta}{1-\tan ^{2} \beta}=\frac{2 \cdot \frac{1}{3}}{1-\frac{1}{9}}=\frac{\frac{2}{3}}{\frac{8}{9}}=\frac{3}{4} \\
\tan (\alpha+2 \beta)=\frac{\tan \alpha+\tan 2 \beta}{1-\tan \alpha \tan 2 \beta} \\
=\frac{\frac{1}{7}+\frac{3}{4}}{1-\frac{1}{7} \cdot \frac{3+2}{4}}=\frac{\frac{42}{28}}{\frac{25}{2}}=1
\end{gathered}
\)

Hint

(d)

\(
\begin{aligned}
&\text { Let } \pi x-\frac{\pi}{4}=\theta \in\left[\frac{-\pi}{4}, \frac{7 \pi}{4}\right] \\
&\text { So, }\left(3-\sin \left(\frac{\pi}{2}+2 \theta\right)\right) \sin \theta \geq \sin (\pi+3 \theta) \\
&\Rightarrow(3-\cos 2 \theta) \sin \theta \geq-\sin 3 \theta \\
&\Rightarrow \sin \theta\left[3-4 \sin ^{2} \theta+3-\cos 2 \theta\right] \geq 0 \\
&\Rightarrow \sin \theta(6-2(1-\cos 2 \theta)-\cos 2 \theta) \geq 0 \\
&\Rightarrow \sin \theta(4+\cos 2 \theta) \geq 0 \\
&\Rightarrow \sin \theta \geq 0 \\
&\Rightarrow \theta \in[0, \pi] \Rightarrow 0 \leq \pi x-\frac{\pi}{4} \leq \pi \\
&\Rightarrow x \in\left[\frac{1}{4}, \quad \frac{5}{4}\right] \\
&\Rightarrow[\alpha, \beta]=\left[\frac{1}{4}, \frac{5}{4}\right] \\
&\therefore \beta-\alpha=\frac{5}{4}-\frac{1}{4}=1
\end{aligned}
\)

Hint

(a) \(\text { Given : } A+B=\pi / 3 \Rightarrow \tan (A+B)=\sqrt{3}\)
\(
\begin{aligned}
&\Rightarrow \frac{\tan A+\tan B}{1-\tan A \tan B}=\sqrt{3} \Rightarrow \frac{\tan A+\frac{y}{\tan A}}{1-y}=\sqrt{3} \\
&\Rightarrow \tan ^{2} A+\sqrt{3}(y-1) \tan A+y=0 \quad \text { [Let } y=\tan A \tan B \text { ] }
\end{aligned}
\)
For real value of \(\tan A, 3(y-1)^{2}-4 y \geq 0\)
\(
\begin{aligned}
&\Rightarrow 3 y^{2}-10 y+3 \geq 0 \Rightarrow(y-3)\left(y-\frac{1}{3}\right) \geq 0 \\
&\Rightarrow y \leq \frac{1}{3} \text { or } \geq 3
\end{aligned}
\)
But \(A, B>0\) and \(A+B=\pi / 3 \Rightarrow A, B<\pi / 3\)
\(
\begin{aligned}
&\Rightarrow \tan A \tan B<3 \\
&\Rightarrow y \leq \frac{1}{3} \Rightarrow \text { Max. value of } y \text { is } 1 / 3 .
\end{aligned}
\)

Hint

(c) \(K=\sin \frac{\pi}{18} \sin \frac{5 \pi}{18} \sin \frac{7 \pi}{18}\) \(=\cos \left(\frac{\pi}{2}-\frac{\pi}{18}\right) \cos \left(\frac{\pi}{2}-\frac{5 \pi}{18}\right) \cos \left(\frac{\pi}{2}-\frac{7 \pi}{18}\right)\) \(=\cos \frac{\pi}{9} \cos \frac{2 \pi}{9} \cos \frac{4 \pi}{9}=\frac{1}{2^{3} \sin \frac{\pi}{9}} \cdot \sin \frac{8 \pi}{9}\) \(\left[\cos \alpha \cos 2 \alpha \quad \cos 2^{2} \alpha \ldots \ldots \ldots \cos 2^{n-1} \alpha=\frac{1}{2^{n} \sin \alpha} \cdot \sin \left(2^{n} \alpha\right)\right]\) \(=\frac{1}{8 \sin \pi / 9} \cdot \sin \pi / 9=\frac{1}{8}\)

Hint

(b)

\(\sin \frac{\pi}{14} \sin \frac{3 \pi}{14} \sin \frac{5 \pi}{14} \sin \frac{7 \pi}{14} \sin \frac{9 \pi}{14} \sin \frac{11 \pi}{14} \sin \frac{13 \pi}{14}\)
\(
\begin{aligned}
=\left(\sin \frac{\pi}{14} \sin \frac{3 \pi}{14} \sin \frac{5 \pi}{14}\right)^{2} \times 1 \\
& {\left[\because \sin \frac{13 \pi}{14}=\sin \frac{\pi}{14}, \sin \frac{11 \pi}{14}=\sin \frac{3 \pi}{14} \text { and } \sin \frac{3 \pi}{14}=\sin \frac{5 \pi}{14}\right] }
\end{aligned}
\)
\(\begin{aligned}
&=\left[\cos \left(\frac{\pi}{2}-\frac{\pi}{14}\right) \cos \left(\frac{\pi}{2}-\frac{3 \pi}{14}\right) \cos \left(\frac{\pi}{2}-\frac{5 \pi}{14}\right)\right]^{2} \\
&=\left[\cos \frac{3 \pi}{7} \cos \frac{2 \pi}{7} \cos \frac{\pi}{7}\right]^{2}=\left[\cos \frac{\pi}{7} \cos \frac{2 \pi}{7} \cos \frac{4 \pi}{7}\right]^{2}
\end{aligned}\)

\(\left[\cos \alpha \cos 2 \alpha \quad \cos 2^{2} \alpha \ldots \ldots \ldots \cos 2^{n-1} \alpha\right.\)\(\left.=\frac{1}{2^{n} \sin \alpha} \cdot \sin \left(2^{n} \alpha\right)\right]\)

\(
\begin{aligned}
&=\left(\frac{1}{8 \sin \pi / 7} \sin \frac{8 \pi}{7}\right)^{2}=\left(\frac{\sin (\pi+\pi / 7)}{8 \sin \pi / 7}\right)^{2} \\
&=\left(\frac{-\sin \pi / 7}{8 \sin \pi / 7}\right)^{2}=\left(\frac{1}{8}\right)^{2}=\frac{1}{64}
\end{aligned}
\)

Hint

(c) Given \(\sin ^{3} x \cdot \sin 3 x=\sum_{m=0}^{n} C_{m} \cos m x\)
\(
\begin{aligned}
&\sin ^{3} x \sin 3 x=\frac{1}{4}[3 \sin x-\sin 3 x] \sin 3 x \\
&=\quad \frac{1}{4}\left[\frac{3}{2} \cdot 2 \sin x \cdot \sin 3 x-\sin ^{2} 3 x\right] \\
&=\frac{1}{4}\left[\frac{3}{2}(\cos 2 x-\cos x)-\frac{1}{2}(1-\cos 6 x)\right] \\
&=\frac{1}{8}[\cos 6 x+3 \cos 2 x-3 \cos x-1] \\
&\quad \therefore \quad \text { Max value of } m=6 \quad \therefore \quad n=6
\end{aligned}
\)

Hint

(b) (True) \(\tan A=\frac{1-\cos B}{\sin B}=\frac{2 \sin ^{2} B / 2}{2 \sin B / 2 \cos / 2}=\tan B / 2\)
Now, \(\tan 2 A=\frac{2 \tan A}{1-\tan ^{2} A}=\frac{2 \tan B / 2}{1-\tan ^{2} B / 2}=\tan B\) Hence, statement is true.

Hint

(b, c) Given: \(f(x)=x \sin \pi x, x>0\) \(\Rightarrow f^{\prime}(x)=\sin \pi x+x \pi \cos \pi x\)
Now, \(f^{\prime}(x)=0 \Rightarrow \tan \pi x=-\pi x\)

From graph of \(y=\tan \pi x\) and \(y=-\pi x\), it is clear that they intersect each other at unique point in the intervals
\((n, n+1) \text { and }\left(n+\frac{1}{2}, n+1\right)\)

Hint

\((a, c, d)\)
As \(\tan (2 \pi-\theta)>0\) and \(-1<\sin \theta<-\frac{\sqrt{3}}{2}, \theta \in[0,2 \pi]\)
Hence \(\frac{3 \pi}{2}<\theta<\frac{5 \pi}{3}\)
Now \(2 \cos \theta(1-\sin \phi)=\sin ^{2} \theta\left(\tan \frac{\theta}{2}+\cot \frac{\theta}{2}\right) \cos \phi-1\)
\(
\Rightarrow 2 \cos \theta(1-\sin \phi)=2 \sin \theta \cos \phi-1
\)
\(
\Rightarrow 2 \cos \theta+1=2 \sin (\theta+\phi)
\)
As \(\theta \in\left(\frac{3 \pi}{2}, \frac{5 \pi}{3}\right), 1<2 \sin (\theta+\phi)<2\)
As \(\theta+\varphi \in\left(\frac{\pi}{6}, \frac{5 \pi}{6}\right)\) or \((\theta+\phi) \in\left(\frac{13 \pi}{6}, \frac{17 \pi}{6}\right)\)
We have \(\varphi \in\left(-\frac{3 \pi}{2},-\frac{2 \pi}{3}\right) \cup\left(\frac{2 \pi}{3}, \frac{7 \pi}{6}\right)\)

Hint

(a, b) Given:
\(
\frac{\sin ^{4} x}{2}+\frac{\cos ^{4} x}{3}=\frac{1}{5} \Rightarrow 3 \sin ^{4} x+2 \cos ^{4} x=\frac{6}{5}
\)
\(
\begin{aligned}
&\Rightarrow \sin ^{4} x+2\left[\sin ^{4} x+\cos ^{4} x\right]=\frac{6}{5} \\
&\Rightarrow \sin ^{4} x+2\left[1-2 \sin ^{2} x \cos ^{2} x\right]=\frac{6}{5} \\
&\Rightarrow \sin ^{4} x+2-4 \sin ^{2} x\left(1-\sin ^{2} x\right)=\frac{6}{5} \\
&\Rightarrow 5 \sin ^{4} x-4 \sin ^{2} x+2-\frac{6}{5}=0 \\
&\Rightarrow 25 \sin ^{4} x-20 \sin ^{2} x+4=0 \\
&\Rightarrow\left(5 \sin ^{2} x-2\right)^{2}=0 \Rightarrow \sin ^{2} x=\frac{2}{5} \\
&\Rightarrow \cos ^{2} x=\frac{3}{5} \text { and } \tan ^{2} x=\frac{2}{3} \\
&\text { Also } \frac{\sin ^{8} x}{8}+\frac{\cos ^{8} x}{27}=\frac{2}{625}+\frac{3}{625}=\frac{5}{625}=\frac{1}{125}
\end{aligned}
\)

Hint

(a, b, c, d) Note that the multiplicative loop is very important approach in IIT Mathematics
\(
\begin{aligned}
&\left(\tan \frac{\theta}{2}\right)(1+\sec \theta)=\frac{\sin \theta / 2}{\cos \theta / 2} \cdot\left[1+\frac{1}{\cos \theta}\right] \\
&=\frac{(\sin \theta / 2) 2 \cos ^{2} \theta / 2}{(\cos \theta / 2) \cos \theta} \\
&=\frac{(2 \sin \theta / 2) \cos \theta / 2}{\cos \theta}=\frac{\sin \theta}{\cos \theta}=\tan \theta \\
&\therefore \quad f_{n}(\theta)=(\tan \theta / 2)(1+\sec \theta) \\
&(1+\sec 2 \theta)\left(1+\sec 2^{2} \theta\right) \ldots\left(1+\sec 2^{n} \theta\right) \\
&=(\tan \theta)(1+\sec 2 \theta)\left(1+\sec 2^{2} \theta\right) \ldots\left(1+\sec 2^{n} \theta\right) \\
&=\tan 2 \theta \cdot\left(1+\sec 2^{2} \theta\right) \ldots\left(1+\sec 2^{n} \theta\right)=\tan \left(2^{n} \theta\right) \\
&\text { Now, } f_{2}\left(\frac{\pi}{16}\right)=\tan \left(2^{2} \cdot \frac{\pi}{16}\right)=\tan \left(\frac{\pi}{4}\right)=1
\end{aligned}
\)
Therefore, (a) is the correct option.
\(
f_{3}\left(\frac{\pi}{32}\right)=\tan \left(2^{3} \cdot \frac{\pi}{32}\right)=\tan \left(\frac{\pi}{4}\right)=1
\)
Therefore, (b) is the correct option.
\(
f_{4}\left(\frac{\pi}{64}\right)=\tan \left(2^{4} \cdot \frac{\pi}{64}\right)=\tan \left(\frac{\pi}{4}\right)=1
\)
Therefore, (c) is the correct option.
\(
f_{5}\left(\frac{\pi}{128}\right)=\tan \left(2^{5} \cdot \frac{\pi}{128}\right)=\tan \left(\frac{\pi}{4}\right)=1
\)
Therefore, (d) is the correct option.

Hint

(c) \(\sin \alpha+\sin \beta+\sin \gamma\)
\(
=2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}+2 \sin \frac{\gamma}{2} \cos \frac{\gamma}{2}
\)
\(
\begin{aligned}
&=2 \sin \left(\frac{\pi}{2}-\frac{\gamma}{2}\right) \cos \frac{\alpha-\beta}{2}+2 \sin \left(\frac{\pi}{2}-\frac{\alpha+\beta}{2}\right) \cos \frac{\gamma}{2} \\
&=2 \cos \frac{\gamma}{2}\left[\cos \frac{\alpha-\beta}{2}+\cos \frac{\alpha+\beta}{2}\right] \\
&=2 \cos (\alpha / 2) \cos (\beta / 2) \cos (\gamma / 2)
\end{aligned}
\)
\(\therefore\) Each \(\cos ((\alpha / 2), \cos (\beta / 2), \cos (\gamma / 2)\) lies between \(-1\) and 1 .
\(
\begin{aligned}
&\Rightarrow-1 \leq \cos \alpha / 2, \cos \beta / 2, \cos \gamma / 2 \leq 1 \\
&\Rightarrow-2 \leq 2 \cos \alpha / 2, \cos \beta / 2, \cos \gamma / 2 \leq 2 \\
&\Rightarrow-2 \leq \cos \alpha+\cos \beta+\cos \gamma \leq 2 \\
&\therefore \quad \text { Min value of } \sin \alpha+\sin \beta+\sin \gamma=-2
\end{aligned}
\)

Hint

(d) \(2 \sin ^{2} x+3 \sin x-2>0\)
\(
\begin{aligned}
&(2 \sin x-1)(\sin x+2)>0 \\
&\quad \Rightarrow 2 \sin x-1>0 \quad(\because-1 \leq \sin x \leq 1) \\
&\quad \Rightarrow \sin x>1 / 2 \Rightarrow x \in(\pi / 6,5 \pi / 6) \\
&\text { Also } x^{2}-x-2<0 \\
&\Rightarrow(x-2)(x+1)<0 \Rightarrow-1<x<2
\end{aligned}
\)
On combining (i) and (ii), we get \(x \in(\pi / 6,2)\).

Hint

(b) \(3\left[\sin ^{4}\left(\frac{3 \pi}{2}-\alpha\right)+\sin ^{4}(3 \pi+\alpha)\right]\)
\(
-2\left[\sin ^{6}(\pi / 2+\alpha)+\sin ^{6}(5 \pi-\alpha)\right]
\)
\(
\begin{aligned}
&=3\left[\cos ^{4} \alpha+\sin ^{4} \alpha\right]-2\left[\cos ^{6} \alpha+\sin ^{6} \alpha\right] \\
&=3\left[\left(\cos ^{2} \alpha+\sin ^{2} \alpha\right)^{2}-2 \sin ^{2} \alpha \cos ^{2} \alpha\right] \\
&-2\left[\left(\cos ^{2} \alpha+\sin ^{2} \alpha\right)^{3}-3 \cos ^{2} \alpha \sin ^{2} \alpha\left(\cos ^{2} \alpha+\sin ^{2} \alpha\right)\right]= \\
&\quad 3\left[1-2 \sin ^{2} \alpha \cos ^{2} \alpha\right]-2\left[1-3 \cos ^{2} \alpha \sin ^{2} \alpha\right] \\
&=3-6 \sin ^{2} \alpha \cos ^{2} \alpha-2+6 \sin ^{2} \alpha \cos ^{2} \alpha=1
\end{aligned}
\)

Hint

(c) Given,
\(
\begin{aligned}
&(1+\cos \pi / 8)(1+\cos 3 \pi / 8)(1+\cos 5 \pi / 8)(1+\cos 7 \pi / 8) \\
=&(1+\cos \pi / 8)(1+\cos 3 \pi / 8)(1+\cos (\pi-3 \pi / 8)) (1+\cos (\pi-\pi / 8)) \\
=&(1+\cos \pi / 8)(1+\cos 3 \pi / 8)(1-\cos 3 \pi / 8)(1-\cos \pi / 8)=\\
&\left(1-\cos ^{2} \pi / 8\right)\left(1-\cos ^{2} 3 \pi / 8\right)=\sin ^{2} \pi / 8 \sin ^{2} 3 \pi / 8 \\
=& \frac{1}{4} \cdot[2 \sin \pi / 8 \sin (\pi / 2-\pi / 8)]^{2} \\
=& \frac{1}{4} \cdot[2 \sin \pi / 8 \cos (\pi / 8)]^2=\frac{1}{4} \cdot \sin ^{2} \pi / 4=\frac{1}{4} \times \frac{1}{2}=\frac{1}{8}
\end{aligned}
\)\(\)

Hint

\(
\begin{aligned}
&\because A+B+C=\pi \\
&\Rightarrow \quad \frac{A}{2}+\frac{B}{2}+\frac{C}{2}=\frac{\pi}{2} \Rightarrow \frac{A}{2}+\frac{B}{2}=\frac{\pi}{2}-\frac{C}{2} \\
&\Rightarrow \cot \left(\frac{A}{2}+\frac{B}{2}\right)=\cot \left(\frac{\pi}{2}-\frac{C}{2}\right) \\
&\Rightarrow \quad \frac{\cot \frac{A}{2} \cdot \cot \frac{B}{2}-1}{\cot \frac{A}{2}+\cot \frac{B}{2}}=\tan \frac{C}{2} \\
&\Rightarrow \cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{C}{2}=\cot \frac{A}{2} \cdot \cot \frac{B}{2} \cdot \cot \frac{C}{2}
\end{aligned}
\)

Hint

\(\text { Let } S=\sum_{k=1}^{n-1}(n-k) \cos \frac{2 k \pi}{n}\)
\(
\begin{aligned}
\Rightarrow S=&(n-1) \cos \frac{2 \pi}{n}+(n-2) \cos 2 . \frac{2 \pi}{n}+\ldots \ldots \\
&+1 \cdot \cos (n-1) \cos \frac{2(\mathrm{n}-1) \pi}{n} \quad \ldots \text { (i) }
\end{aligned}
\)
We know that \(\cos \theta=\cos (2 \pi-\theta)\)
Replacing each angle \(\theta\) by \(2 \pi-\theta\) in (i), we get
\(
\begin{array}{r}
S=(n-1) \cos (n-1) \frac{2 \pi}{n}+(n-2) \cos (n-2) \frac{2 \pi}{n}+\ldots \ldots \\
+1 . \cos \frac{2 \pi}{n} \ldots \text { (ii) }
\end{array}
\)
On adding terms in (i) and (ii) having the same angle and taking \(n\) common, we get
\(
\therefore 2 S=n\left[\cos \frac{2 \pi}{n}+\cos \frac{4 \pi}{n}+\cos \frac{6 \pi}{n}+\ldots .+\cos (n-1) \frac{2 \pi}{n}\right]
\)
Angles are in A.P. with common difference \((d)=\frac{2 \pi}{n}\)
\(
\begin{aligned}
2 S &=n\left[\frac{\sin (n-1) \frac{\pi}{n}}{\sin \frac{\pi}{n}} \cos \frac{\frac{2 \pi}{n}+(n-1) \frac{2 \pi}{n}}{2}\right] \\
&=n .1 \cos \pi=-n, \quad \because \sin (\pi-\theta)=\sin \theta, \quad \therefore S=-n / 2
\end{aligned}
\)

Hint

Let \(y=\frac{\sin x \cos 3 x}{\sin 3 x \cos x}=\frac{\tan x}{\tan 3 x}\)
\(
\begin{aligned}
&=\frac{\tan x\left(1-3 \tan ^{2} x\right)}{3 \tan x-\tan ^{3} x}=\frac{1-3 \tan ^{2} x}{3-\tan ^{2} x} \\
\Rightarrow & 3 y-\left(\tan ^{2} x\right) y=1-3 \tan ^{2} x \Rightarrow 3 y-1=(y-3) \tan ^{2} x \\
\Rightarrow & \tan ^{2} x=\frac{3 y-1}{y-3}=\frac{(3 y-1)(y-3)}{(y-3)^{2}} \\
\text { Since, } \tan ^{2} x>0, \quad \therefore(3 y-1)(y-3)>0 \\
\Rightarrow &\left(y-\frac{1}{3}\right)(y-3)>0 \Rightarrow y<\frac{1}{3} \quad \text { or } \quad y>3
\end{aligned}
\)
\(
\therefore y \text { cannot lie between } \frac{1}{3} \text { and } 3 \text {. }
\)

Hint

Given: \(\cos \theta=\sin \varphi\), where \(\theta=p \sin x, \varphi=p \cos x\)
Above is possible when both \(\theta=\varphi=\frac{\pi}{4}\) or \(\theta=\varphi=\frac{5 \pi}{4}\)
\(
\therefore \quad p \sin x=\frac{\pi}{4} \quad \text { or } \quad p \sin x=\frac{5 \pi}{4}
\)
and \(p \cos x=\frac{\pi}{4}\) or \(p \cos x=\frac{5 \pi}{4}\)
On squaring and adding, \(p^{2}=\frac{\pi^{2}}{16} .2\) or \(\frac{25 \pi^{2}}{16} .2\)
\(\therefore \quad p=\frac{\pi}{4} \sqrt{2}\) only for least positive value or \(p=\frac{\pi}{4} \sqrt{2}\)

Hint

Given: \(\tan \left(x+100^{\circ}\right)=\tan \left(x+50^{\circ}\right) \tan x \tan \left(x-50^{\circ}\right)\)
\(
\begin{aligned}
&\Rightarrow \frac{\tan \left(x+100^{\circ}\right)}{\tan x}=\tan \left(x+50^{\circ}\right) \tan \left(x-50^{\circ}\right) \\
&\Rightarrow \frac{\sin \left(x+100^{\circ}\right) \cos x}{\cos \left(x+100^{\circ}\right) \sin x}=\frac{\sin \left(x+50^{\circ}\right) \sin \left(x-50^{\circ}\right)}{\cos \left(x+50^{\circ}\right) \cos \left(x-50^{\circ}\right)} \\
&\Rightarrow \frac{\sin \left(2 x+100^{\circ}\right)+\sin 100^{\circ}}{\sin \left(2 x+100^{\circ}\right)-\sin 100^{\circ}}=\frac{\cos 100^{\circ}-\cos 2 x}{\cos 100^{\circ}+\cos 2 x}
\end{aligned}
\)
By componendo and dividendo,
\(/
\begin{aligned}
&\Rightarrow \frac{2 \sin \left(2 x+100^{\circ}\right)}{2 \sin 100^{\circ}}=\frac{2 \cos 100^{\circ}}{-2 \cos 2 x} \\
&\Rightarrow 2 \sin \left(2 x+100^{\circ}\right) \cos 2 x=-2 \sin 100^{\circ} \cos 100^{\circ} \\
&\Rightarrow \sin \left(4 x+100^{\circ}\right)+\sin 100^{\circ}=-\sin 200^{\circ} \\
&\Rightarrow \sin \left(4 x+10^{\circ}+90^{\circ}\right)+\sin \left(90^{\circ}+10^{\circ}\right)=-\sin \left(180+20^{\circ}\right) \\
&\Rightarrow \cos \left(4 x+10^{\circ}\right)+\cos 10^{\circ}=\sin 20^{\circ} \\
&\Rightarrow \cos \left(4 x+10^{\circ}\right)=\sin 20^{\circ}-\cos 10^{\circ} \\
&\Rightarrow \cos \left(4 x+10^{\circ}\right)=\sin 20^{\circ}-\sin 80^{\circ} \\
&=-2 \cos 50^{\circ} \sin 30^{\circ}=-2 \cos 50^{\circ} \cdot \frac{1}{2} \\
&=-\cos 50^{\circ}=\cos 130^{\circ}
\end{aligned}
\)
\(\Rightarrow 4 x+10^{\circ}=130^{\circ} \Rightarrow x=30^{\circ}\)

Hint

Let \(y=\frac{\tan x}{\tan 3 x} \Rightarrow y=\frac{\tan x\left(1-3 \tan ^{2} x\right)}{3 \tan x-\tan ^{3} x}\) \(\Rightarrow 3 y-3 \tan ^{2} x=1-3 \tan ^{2} x\)
\(\Rightarrow(y-3) \tan ^{2} x=3 y-1 \Rightarrow \tan ^{2} x=\frac{3 y-1}{y-3}\)
\(\Rightarrow \frac{3 y-1}{y-3}>0 \quad(\because\) L.H.S. is a prefect square \()\)
\(\Rightarrow \frac{(3 y-1)(y-3)}{(y-3)^{2}}>0 \Rightarrow(3 y-1)(y-3)>0\)
\(\Rightarrow y<\frac{1}{3}\) or \(y>3\)
Thus \(y\) never lies between \(\frac{1}{3}\) and 3 .

Hint

Let \(y=\exp \left[\sin ^{2} x+\sin ^{4} x+\sin ^{6} x+\ldots \infty\right] \ln 2\)
\(
\begin{aligned}
&=e^{I n \cdot 2^{\sin ^{2} x+\sin ^{4} x+\sin ^{6} x+\ldots \infty}} \\
&=2^{\sin ^{2} x+\sin ^{4} x+\sin ^{6} x+\ldots . \infty}=\frac{\sin ^{2} x}{2^{1-\sin ^{2} x}}=2^{\tan ^{2} x}
\end{aligned}
\)
As \(y\) satisfies the eq. \(x^{2}-9 x+8=0\)
\(\therefore y^{2}-9 y+8=0\)
\(
\begin{aligned}
&\Rightarrow(y-1)(y-8)=0 \Rightarrow y=1,8 \\
&\Rightarrow 2^{\tan ^{2} x}=1 \text { or } 2^{\tan ^{2} x}=8 \\
&\Rightarrow \tan ^{2} x=0 \text { or } \tan ^{2} x=3 \\
&\Rightarrow \tan x=0 \text { or } \tan x=\sqrt{3},-\sqrt{3} \\
&\Rightarrow x=0 \text { or } x=\pi / 3,2 \pi / 3
\end{aligned}
\)
But given \(0<x<\pi / 2 \Rightarrow x=\pi / 3\)
\(\therefore \frac{\cos x}{\cos x+\sin x}=\frac{1}{1+\tan x}=\frac{1}{1+\sqrt{3}}=\frac{\sqrt{3}-1}{2}\)

Hint

Given : In \(\triangle A B C, A, B\) and \(C\) are in A.P.
\(\therefore A+C=2 B\)
\(\mathrm{Also} A+B+C=180^{\circ} \Rightarrow B+2 B=180^{\circ} \Rightarrow B=60^{\circ}\)
Also given that, \(\sin (2 A+B)=\sin (C-A)=-\sin (B+2 C)=\frac{1}{2}\)
\(
\Rightarrow \quad \sin \left(2 A+60^{\circ}\right)=\sin (C-A)=-\sin (60+2 C)=\frac{1}{2} \ldots \text { (i) }
\)
From eq. (i), \(\sin \left(2 A+60^{\circ}\right)=\frac{1}{2} \Rightarrow 2 A+60^{\circ}=30^{\circ}, 150^{\circ}\)
But \(A\) can not be -ve
\(
\therefore 2 A+60^{\circ}=150^{\circ} \Rightarrow 2 A=90^{\circ} \Rightarrow A=45^{\circ}
\)
Again from (i), \(\sin \left(60^{\circ}+2 C\right)=-\frac{1}{2}\)
\(
\begin{aligned}
&\Rightarrow \quad 60^{\circ}+2 C=210^{\circ} \text { or } 330^{\circ} \\
&\Rightarrow C=75^{\circ} \text { or } 135^{\circ}
\end{aligned}
\)
Also from (i), \(\sin (C-A)=\frac{1}{2} \Rightarrow C-A=30^{\circ}, 150^{\circ}\)
For \(A=45^{\circ} ; C=75^{\circ}, 195^{\circ}\)
But \(\mathrm{C}=195^{\circ}\) is not possible.
\(\begin{aligned}
&\therefore C=75^{\circ} \\
&\therefore A=45^{\circ}, B=60^{\circ}, C=75^{\circ} .
\end{aligned}\)

Hint

We know \(\tan 2 \alpha=\frac{2 \tan \alpha}{1-\tan ^{2} \alpha}\) \(\Rightarrow \frac{1-\tan ^{2} \alpha}{\tan \alpha}=2 \cot 2 \alpha \Rightarrow \cot \alpha-\tan \alpha=2 \cot 2 \alpha \dots …(i)\)
Now, we have to prove \(\tan \alpha+2 \tan 2 \alpha+4 \tan 4 \alpha+8 \cot 8 \alpha=\cot \alpha\)
LHS \(=\tan \alpha+2 \tan 2 \alpha+4 \tan 4 \alpha+4(2 \cot 2.4 \alpha)\) \(=\tan \alpha+2 \tan 2 \alpha+4 \tan 4 \alpha+4(\cot 4 \alpha-\tan 4 \alpha)\) [From (i)]
\(=\tan \alpha+2 \tan 2 \alpha+4 \tan 4 \alpha+4 \cot 4 \alpha-4 \tan 4 \alpha\)
\(=\tan \alpha+2 \tan 2 \alpha+2(2 \cot 2.2 \alpha)\)
\(=\tan \alpha+2 \tan 2 \alpha+2(\cot \alpha-\tan 2 \alpha)\)
\(=\tan \alpha+2 \tan 2 \alpha+2(2 \cot 2 \alpha-\tan 2 \alpha) \quad\) [From (i)]
\(=\tan \alpha+2 \cot 2 \alpha\)
\(=\tan \alpha+(\cot \alpha-\tan \alpha) \quad\) [From (i)]
\(=\cot \alpha=\) RHS.

Hint

We know,
\(
\begin{aligned}
&\cos A \cos 2 A \cos 4 A \ldots \cos 2^{n} A=\frac{1}{2^{n+1} \sin A} \sin \cdot\left(2^{n+1} A\right) \\
&\therefore 16 \cos \frac{2 \pi}{15} \cos 2\left(\frac{2 \pi}{15}\right) \cos 2^{2}\left(\frac{2 \pi}{15}\right) \cos 2^{3}\left(\frac{2 \pi}{15}\right) \\
&=16 \cdot \frac{\sin \left(2^{4} A\right)}{2^{4} \sin A}, \text { where } A=2 \pi / 15 \\
&=16 . \frac{\sin (32 \pi / 15)}{16 \sin 2 \pi / 15}=\frac{\sin (32 \pi / 15)}{\sin (2 \pi+2 \pi / 15)}=\frac{\sin (32 \pi / 15)}{\sin (32 \pi / 15)}=1
\end{aligned}
\)

Hint

(b)

\(
\begin{aligned}
& =\frac{1}{2}\left(2 \sin 12^{\circ} \sin 48^{\circ}\right) \sin 54^{\circ} \\
& =\frac{1}{2}\left[\cos \left(36^{\circ}\right)-\cos \left(60^{\circ}\right)\right] \sin 54^{\circ} \\
& =\frac{1}{2}\left(\cos 36^{\circ}-\frac{1}{2}\right) \sin 54^{\circ} \\
& =\frac{1}{4}\left(2 \cos 36^{\circ} \sin 54^{\circ}-\sin 54^{\circ}\right) \\
& =\frac{1}{4}\left(\sin 90^{\circ}+\sin 18^{\circ}-\sin 54^{\circ}\right) \\
& =\frac{1}{4}\left(1+\frac{\sqrt{5}-1}{4}-\frac{\sqrt{5}+1}{4}\right) \\
& =\frac{1}{4}\left(1+\frac{\sqrt{5}-1-\sqrt{5}-1}{4}\right) \\
& =\frac{1}{4}\left(1-\frac{1}{2}\right)=\frac{1}{8}
\end{aligned}
\)

Hint

\(\begin{aligned}
&\text {We know, }\\
&\cos \theta+\sin \theta=\sqrt{2}\left[\frac{1}{\sqrt{2}} \cos \theta+\frac{1}{\sqrt{2}} \sin \theta\right]\\
&=\sqrt{2} \sin (\pi / 4+\theta)\\
&\therefore \cos \theta+\sin \theta \leq \sqrt{2}<\pi / 2\left[\begin{array}{c}
\because \sqrt{2}=1.414 \\
\pi / 2=1.57
\end{array}\right]\\
&\therefore \quad \cos \theta+\sin \theta<\pi / 2 \quad \Rightarrow \quad \cos \theta<\pi / 2-\sin \theta \quad \ldots \text { (i) } \quad \text { As }\\
&\theta \in[0, \pi / 2] \text { in which } \sin \theta \text { increases. }\\
&\therefore \text { Taking sin on both sides of eq. (i), we get }\\
&\sin (\cos \theta)<\sin (\pi / 2-\sin \theta)\\
&\sin (\cos \theta)<\cos (\sin \theta)\\
&\Rightarrow \cos (\sin \theta)>\sin (\cos \theta)
\end{aligned}\)

Hint

\(
\begin{aligned}
&A=\left\{x: \frac{\pi}{6} \leq x \leq \frac{\pi}{3}\right\} \\
&f(x)=\cos x-x(1+x) \\
&f^{\prime}(x)=-\sin x-1-2 x<0, \forall x \in A
\end{aligned}
\)
\(\therefore f\) is a decreasing function.
As
\(
\begin{aligned}
\frac{\pi}{6} & \leq x \leq \frac{\pi}{3} \Rightarrow f\left(\frac{\pi}{3}\right) \leq f(x) \leq\left(\frac{\pi}{6}\right) \\
& \Rightarrow \cos \frac{\pi}{3}-\frac{\pi}{3}\left(1+\frac{\pi}{3}\right) \leq f(x) \leq \cos \frac{\pi}{6}-\frac{\pi}{6}\left(1+\frac{\pi}{6}\right) \\
\therefore & f(A)=\left[\frac{1}{2}-\frac{\pi}{3}\left(1+\frac{\pi}{3}\right), \frac{\sqrt{3}}{2}-\frac{\pi}{6}\left(1+\frac{\pi}{6}\right)\right]
\end{aligned}
\)

Hint

Given \(\alpha+\beta-\gamma=\pi\) and we have to prove that
\(\sin ^{2} \alpha+\sin ^{2} \beta-\sin ^{2} \gamma=2 \sin \alpha \sin \beta \cos \gamma\)
L.H.S. \(=\sin ^{2} \alpha+\sin ^{2} \beta-\sin ^{2} \gamma\)
\(=\sin ^{2} \alpha+\sin (\beta+\gamma) \sin (\beta-\gamma)\)
\(=\sin ^{2} \alpha+\sin (\beta+\gamma) \sin (\pi-\alpha) \quad\left[\text { Since }, \sin ^{2} A-\sin ^{2} B=\sin (A+B) \sin (A-B)\right]\)
\(=\sin ^{2} \alpha+\sin (\beta+\gamma) \sin (\pi-\gamma)\) \(=\sin ^{2} \alpha+\sin (\beta+\gamma) \sin \alpha\)
\(=\sin \alpha(\sin \alpha+\sin (\beta+\gamma))\)
\(=\sin \alpha[\sin [\pi-(\beta-\gamma)]+\sin (\beta+\gamma)]\)
\(=\sin \alpha[\sin (\beta-\gamma)+\sin (\beta+\gamma)]\)
\(=\sin \alpha[2 \sin \beta \cos \gamma]=2 \sin \alpha \sin \beta \cos \gamma=\) R.H.S.

Hint

\(
\begin{aligned}
&\text { 52. We know } \tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta} \\
&\Rightarrow \tan (\alpha+\beta)=\frac{\frac{m}{m+1}+\frac{1}{2 m+1}}{1-\frac{m}{m+1} \cdot \frac{1}{2 m+1}}=\frac{2 m^{2}+2 m+1}{2 m^{2}+2 m+1}=1 \\
&\Rightarrow \alpha+\beta=n \pi+\pi / 4, \text { where } n \in Z
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (b) } \sin ^{4} \theta+\cos ^{4} \theta=-\lambda\\
&\Rightarrow\left(\sin ^{2} \theta+\cos ^{2} \theta\right)^{2}-2 \sin ^{2} \theta \cdot \cos ^{2} \theta=-\lambda\\
&\Rightarrow 1-2 \sin ^{2} \theta \cos ^{2} \theta=-\lambda\\
&\Rightarrow \lambda=\frac{(\sin 2 \theta)^{2}}{2}-1\\
&\Rightarrow \text { as } \sin ^{2} 2 \theta \in[0,1]
\end{aligned}
\)

\(
\Rightarrow \lambda \in\left[-1, \frac{-1}{2}\right]
\)

Hint

(a) According to the question, there are two cases.
Case \(1: \theta \in\left(\frac{\pi}{2}, \frac{2 \pi}{3}\right)\)
In this interval, \([\sin \theta]=0,[-\cos \theta]=0\) and \([\cot \theta]=-1\)
Then the system of equations will be ;
\(0 \cdot x+0 \cdot y=0\) and \(-x+y=0\)
Which have infinitely many solutions.
Case 2: \(\theta \in\left(\pi, \frac{7 \pi}{6}\right)\)
In this interval, \([\sin \theta]=-1\) and \([-\cos \theta]=0\),
Then the system of equations will be;
\(-x+0 \cdot y=0\) and \([\cot \theta] x+y=0\)
Clearly, \(x=0\) and \(y=0\) which has unique solution.

Hint

(c) \(2 \cos ^{2} \theta+3 \sin \theta=0\)
\((2 \sin \theta+1)(\sin \theta-2)=0\)
\(\Rightarrow \sin \theta=-\frac{1}{2}\) or \(\sin \theta=2 \rightarrow\) Not possible
The required sum of all solutions in \([-2 \pi, 2 \pi]\) is
\(
=\left(\pi+\frac{\pi}{6}\right)+\left(2 \pi-\frac{\pi}{6}\right)+\left(-\frac{\pi}{6}\right)+\left(-\pi+\frac{\pi}{6}\right)=2 \pi
\)

Hint

(d)

\(
\begin{aligned}
&\sin x+\sin 3 x=\sin 2 x \Rightarrow 2 \sin 2 x \cos x=\sin 2 x \\
&\Rightarrow \sin 2 x=0 \text { or } \cos x=1 / 2 \\
&\Rightarrow 2 x \in\{0\}, \quad x \in\{\pi / 3\} \\
&\Rightarrow x={0, \frac{\pi}{3}}, \quad x \in\left[0, \frac{\pi}{2}\right]
\end{aligned}
\)

Hint

(a) \(8 \cos x\left(\cos ^{2} \frac{\pi}{6}-\sin ^{2} x-\frac{1}{2}\right)=1\)
\(\Rightarrow \quad 8 \cos x\left(\frac{3}{4}-\frac{1}{2}-\sin ^{2} x\right)=1\)
\(\Rightarrow \quad 8 \cos x\left(\frac{1}{4}-\left(1-\cos ^{2} x\right)\right)=1\)
\(\Rightarrow \quad 8 \cos x\left(\frac{1}{4}-1+\cos ^{2} x\right)=1\)
\(\Rightarrow 8 \cos x\left(\cos ^{2} x-\frac{3}{4}\right)=1\)
\(\Rightarrow 8\left(\frac{4 \cos ^{3} x-3 \cos x}{4}\right)=1\)
\(\Rightarrow 2\left(4 \cos ^{3} x-3 \cos x\right)=1\)
\(\Rightarrow 2 \cos 3 x=1 \Rightarrow \cos 3 x=\frac{1}{2}\)
\(\therefore 3 \mathrm{x}=2 \mathrm{n} \pi \pm \frac{\pi}{3}, \mathrm{n} \in 1\)
\(\Rightarrow \mathrm{x}=\frac{2 \mathrm{n} \pi}{3} \pm \frac{\pi}{9}\)
In \(x \in[0, \pi]: x=\frac{\pi}{9}, \frac{2 \pi}{3}+\frac{\pi}{9}, \frac{2 \pi}{3}-\frac{\pi}{9}\), only Sum of all the solutions of the equation \(=\left(\frac{1}{9}+\frac{2}{3}+\frac{1}{9}+\frac{2}{3}-\frac{1}{9}\right) \pi=\frac{13}{9} \pi\)

Hint

\(
\begin{aligned}
&\text { (a) } \cos x+\cos 2 x+\cos 3 x+\cos 4 x=0\\
&\Rightarrow 2 \cos 2 \mathrm{x} \cos \mathrm{x}+2 \cos 3 \mathrm{x} \cos \mathrm{x}=0\\
&\Rightarrow 2 \cos x\left(2 \cos \frac{5 x}{2} \cos \frac{x}{2}\right)=0\\
&\cos x=0, \cos \frac{5 x}{2}=0, \cos \frac{x}{2}=0\\
&\mathrm{x}=\pi, \frac{\pi}{2}, \frac{3 \pi}{2}, \frac{\pi}{5}, \frac{3 \pi}{5}, \frac{7 \pi}{5}, \frac{9 \pi}{5}
\end{aligned}
\)

Hint

(d) \(\left|\sqrt{2 \sin ^{4} x+18 \cos ^{2} x}-\sqrt{2 \cos ^{4} x+18 \sin ^{2} x}\right|=1\)
\(\sqrt{2 \sin ^{4} x+18 \cos ^{2} x}-\sqrt{2 \cos ^{4} x+18 \sin ^{2} x}=\pm 1\)
\(\sqrt{2 \sin ^{4} x+18 \cos ^{2} x}=\pm 1+\sqrt{2 \cos ^{4} x+18 \sin ^{2} x}\)
by squaring both the sides we will get 8 solutions

Hint

(c) \(\sqrt{3} \sec x+\operatorname{cosec} x+2(\tan x-\cot x)=0\)
\(
\begin{aligned}
&\Rightarrow \frac{\sqrt{3}}{2} \sin \mathrm{x}+\frac{1}{2} \cos \mathrm{x}=\cos ^{2} \mathrm{x}-\sin ^{2} \mathrm{x} \\
&\Rightarrow \cos \left(\mathrm{x}-\frac{\pi}{3}\right)=\cos 2 \mathrm{x} \Rightarrow \mathrm{x}-\frac{\pi}{3}=2 \mathrm{n} \pi \pm 2 \mathrm{x} \\
&\Rightarrow \mathrm{x}=\frac{2 \mathrm{n} \pi}{3}+\frac{\pi}{9} \text { or } \mathrm{x}=-2 \mathrm{n} \pi-\frac{\pi}{3}
\end{aligned}
\)
For \(\mathrm{x} \in \mathrm{S}, \mathrm{n}=0 \Rightarrow \mathrm{x}=\frac{\pi}{9},-\frac{\pi}{3}\)
Now, \(\mathrm{n}=1 \Rightarrow \mathrm{x}=\frac{7 \pi}{9} ; \quad\) and \(\mathrm{n}=-1 \Rightarrow \mathrm{x}=\frac{-5 \pi}{9}\)
Hence, sum of all values of \(x=\frac{\pi}{9}-\frac{\pi}{3}+\frac{7 \pi}{9}-\frac{5 \pi}{9}=0\)

Hint

\(
\begin{aligned}
&\text { (c) } 2 \sin ^{3} \alpha-7 \sin ^{2} \alpha+7 \sin \alpha-2=0 \\
&\Rightarrow 2 \sin ^{2} \alpha(\sin \alpha-1)-5 \sin \alpha(\sin \alpha-1) +2(\sin \alpha-1)=0 \\
&\Rightarrow(\sin \alpha-1)\left(2 \sin ^{2} \alpha-5 \sin \alpha+2\right)=0 \\
&\Rightarrow \sin \alpha-1=0 \text { or } 2 \sin ^{2} \alpha-5 \sin \alpha+2=0
\end{aligned}
\)

\(\sin \alpha=1\) or \(\sin \alpha=\frac{5 \pm \sqrt{25-16}}{4}=\frac{5 \pm 3}{4}\) \(\alpha=\frac{\pi}{2}\) or \(\sin \alpha=\frac{1}{2}, 2\)
Now, \(\sin \alpha \neq 2\)
for, \(\sin \alpha=\frac{1}{2}\)
\(
\alpha=\frac{\pi}{3}, \frac{2 \pi}{3}
\)
There are three values of \(\alpha\) between \([0,2 \pi]\)

Hint

(d) \(\sin x+2 \sin 2 x-\sin 3 x=3\)
\(\Rightarrow \quad \sin x+4 \sin x \cos x-3 \sin x+4 \sin ^{3} x=3\)
\(\Rightarrow \quad \sin x\left(-2+2 \cos x+4 \sin ^{2} x\right)=3\)
\(\Rightarrow \quad \sin x\left(-2+2 \cos x+4-4 \cos ^{2} x\right)=3\)
\(\Rightarrow \quad 2+2 \cos x-4 \cos ^{2} x=\frac{3}{\sin x}\)
\(\Rightarrow \quad 2-\left(4 \cos ^{2} x-2.2 \cos x \cdot \frac{1}{2}+\frac{1}{4}\right)+\frac{1}{4}=\frac{3}{\sin x}\)
\(\Rightarrow \frac{9}{4}-\left(2 \cos x-\frac{1}{2}\right)^{2}=\frac{3}{\sin x}\)
\(\therefore\) LHS \(\leq \frac{9}{4}\) and RHS \(\geq 3 \quad[\because 0 \leq \sin x \leq 1]\)

Hint

(c) \(2 \sin ^{2} \theta-\cos 2 \theta=0 \Rightarrow 1-2 \cos 2 \theta=0\) \(\Rightarrow \cos 2 \theta=\frac{1}{2} \Rightarrow 2 \theta=\frac{\pi}{3}, \frac{5 \pi}{3}, \frac{7 \pi}{3}, \frac{11 \pi}{3}\) \(\Rightarrow \theta=\frac{\pi}{6}, \frac{5 \pi}{6}, \frac{7 \pi}{6}, \frac{11 \pi}{6} \dots…(i)\)
where \(\theta \in[0,2 \pi]\)
Also,
\(
\begin{aligned}
& 2 \cos ^{2} \theta-3 \sin \theta=0 \\
\Rightarrow & 2 \sin ^{2} \theta+3 \sin \theta-2=0
\end{aligned}
\)
\(
\begin{aligned}
&\Rightarrow(2 \sin \theta-1)(\sin \theta+2)=0 \Rightarrow \sin \theta=\frac{1}{2} \quad[\because \sin \theta \neq-2] \\
&\Rightarrow \theta=\frac{\pi}{6}, \frac{5 \pi}{6} \quad \ldots . \text { (ii), }
\end{aligned}
\)
where \(\theta \in[0,2 \pi]\)
Combining (i) and (ii), we get \(\theta=\frac{\pi}{6}, \frac{5 \pi}{6}\)
Hence, there are two solutions.

Hint

(d) Given : \(\cos (\alpha-\beta)=1\) and \(\cos (\alpha+\beta)=1 / \mathrm{e}\),
where \(\alpha, \beta \in[-\pi, \pi]\)
Now, \(\cos (\alpha-\beta)=1 \Rightarrow \alpha-\beta=0 \Rightarrow \alpha=\beta\)
and \(\cos (\alpha+\beta)=1 / \mathrm{e} \Rightarrow \cos 2 \alpha=1 / \mathrm{e}\)
\(\because 0<1 / e<1\)
Now \(2 \alpha \in[-2 \pi, 2 \pi]\)
\(\Rightarrow\) There will be two values of \(2 \alpha\) satisfying \(\cos 2 \alpha=1 / \mathrm{e}\) and two values in \([0,2 \pi]\) in \([-2 \pi, 0]\).
\(\Rightarrow\) There will be four values of \(\alpha\) in \([-\pi, \pi]\) and correspondingly four values of \(\beta\). Hence there are four sets of \((\alpha, \beta)\).

Hint

(b) We know, \(-\sqrt{a^{2}+b^{2}} \leq a \cos \theta+b \sin \theta \leq \sqrt{a^{2}+b^{2}}\)
\(\Rightarrow-\sqrt{74} \leq 7 \cos x+5 \sin x \leq \sqrt{74}\)
\(\Rightarrow-\sqrt{74} \leq 2 k+1 \leq \sqrt{74} \Rightarrow-8.6 \leq 2 k+1 \leq 8.6\)
\(\Rightarrow-4.8 \leq k \leq 3.8\)
Hence, \(k\) can take only 8 integral values.

Hint

(c) To simplify the det. let \(\sin x=a ; \cos x=b\), then equation becomes
\(
\left|\begin{array}{lll}
a & b & b \\
b & a & b \\
b & b & a
\end{array}\right|=0
\)
\(
\begin{aligned}
&\left|\begin{array}{ccc}
a & b-a & 0 \\
b & a-b & b-a \\
b & 0 & a-b
\end{array}\right|=0\left[C_{2}-C_{1} ; C_{3}-C_{2}\right] \\
&\Rightarrow a(a-b)^{2}-(b-a)[b(a-b)-b(b-a)]=0 \\
&\Rightarrow a(a-b)^{2}-2 b(b-a)(a-b)=0 \\
&\Rightarrow(a-b)^{2}(a-2 b)=0 \Rightarrow(a=b) \text { or } a=2 b \\
&\therefore \frac{a}{b}=1 \text { or } \frac{a}{b}=2 \\
&\Rightarrow \tan x=1 \text { or } \tan x=2
\end{aligned}
\)
But we have \(-\pi / 4 \leq x \leq \pi / 4\)
\(
\begin{aligned}
&\Rightarrow \tan (-\pi / 4) \leq \tan x \leq \tan (\pi / 4) \Rightarrow-1 \leq \tan x \leq 1 \\
&\therefore \tan x=1 \Rightarrow x=\pi / 4
\end{aligned}
\)
Hence, there is only one real not.

Hint

(a) Given : In \(\triangle P Q R, \angle R=\pi / 2\)
\(
\Rightarrow \angle P+\angle Q=\pi / 2 \Rightarrow \frac{\angle P}{2}+\frac{\angle Q}{2}=\frac{\pi}{4}
\)
Also \(\tan P / 2\) and \(\tan Q / 2\) are roots of the equation \(a x^{2}+b x+c=0(a \neq 0)\)
\(
\therefore \tan P / 2+\tan Q / 2=-\frac{b}{a} ; \tan P / 2 \tan Q / 2=c / a
\)
Now \(\tan \left(\frac{P+Q}{2}\right)=\frac{\tan P / 2+\tan Q / 2}{1-\tan P / 2 \tan Q / 2}\)
\(
\begin{aligned}
&\Rightarrow \tan \frac{\pi}{4}=\frac{-b / a}{1-c / a} \Rightarrow 1-\frac{c}{a}=-\frac{b}{a} \\
&\Rightarrow a-c=-b \Rightarrow a+b=c
\end{aligned}
\)

Hint

(b) Given: \(\sec ^{2} \theta=\frac{4 x y}{(x+y)^{2}}\)
But \(\sec ^{2} \theta \geq 1 \Rightarrow \frac{4 x y}{(x+y)^{2}} \geq 1\)
\(
\Rightarrow 4 x y \geq x^{2}+y^{2}+2 x y
\)
\(
\Rightarrow x^{2}+y^{2}-2 x y \leq 0 \Rightarrow(x-y)^{2} \leq 0
\)
\(\Rightarrow x=y\), because perfect square of real number can not be negative.
Also \(x \neq 0\), otherwise \(\sec ^{2} \theta\) will become indeterminate.

Hint

\(
\begin{aligned}
&\text { (d) } 2 \sin ^{2} \theta-3 \sin \theta-2=0 \\
&\Rightarrow(2 \sin \theta+1)(\sin \theta-2)=0 \\
&\Rightarrow \sin \theta=-\frac{1}{2} \quad[\because \sin \theta-2=0, \text { is not possible }] \\
&\Rightarrow \quad \sin \theta=\sin (-\pi / 6) \text { and } \sin (7 \pi / 6) \\
&\Rightarrow \quad \theta=n \pi+(-1)^{n}(-\pi / 6) \text { and } n \pi+(-1)^{n} 7 \pi / 6 \\
&\Rightarrow \text { Thus, } \theta=n \pi+(-1)^{n} 7 \pi / 6
\end{aligned}
\)

Hint

(d) \(\sin \frac{\pi}{2 n}+\cos \frac{\pi}{2 n}=\frac{\sqrt{n}}{2}\)
\(\Rightarrow \sin ^{2} \frac{\pi}{2 n}+\cos ^{2} \frac{\pi}{2 n}+2 \sin \frac{\pi}{2 n} \cos \frac{n}{2 n}=\frac{n}{4}\)
\(\Rightarrow 1+\sin \frac{\pi}{n}=\frac{n}{4} \Rightarrow \sin \frac{\pi}{n}=\frac{n-4}{4}\)
For \(n=2\) the given equation is not satisfied.
Considering \(n>1\) and \(n \neq 2,0<\sin \frac{\pi}{n}<1\)
\(
\Rightarrow 0<\frac{n-4}{4}<1 \Rightarrow 4<n<8 .
\)

Hint

\(
\begin{aligned}
&\text { (c) } \tan x+\sec x=2 \cos x\\
&\Rightarrow \frac{\sin x}{\cos x}+\frac{1}{\cos x}=2 \cos x\\
&\Rightarrow \sin x+1=2 \cos ^{2} x \Rightarrow 2 \sin ^{2} x+\sin x-1=0\\
&\Rightarrow(2 \sin x-1)(\sin x+1)=0 \Rightarrow \sin x=\frac{1}{2},-1\\
&\Rightarrow x=\frac{\pi}{6}, \frac{5 \pi}{6}, \frac{3 \pi}{2} \in[0,2 \pi]
\end{aligned}
\)

But for \(x=\frac{3 \pi}{2}\), given eq. is not defined, Hence, there are only 2 solutions.

Hint

(d) \((\cos p-1) x^{2}+(\cos p) x+\sin p=0\)
For real roots \(D \geq 0\)
\(
\begin{aligned}
& \Rightarrow \cos ^{2} p-4 \sin p(\cos p-1) \geq 0 \\
& \Rightarrow \cos ^{2} p-4 \sin p \cos p+4 \sin ^{2} p+4 \sin p-4 \sin ^{2} p \geq 0 \\
& \Rightarrow(\cos p-2 \sin p)^{2}+4 \sin p(1-\sin p) \geq 0 \\
\text { Since, } p &(\cos p-2 \sin p)^{2} \geq 0 \text { and } 1-\sin p \geq 0 \\
\therefore D \geq 0, & \forall p \in(0, \pi)
\end{aligned}
\)

Hint

(b) \(\sin x-3 \sin 2 x+\sin 3 x=\cos x-3 \cos 2 x+\cos 3 x\)
\(\Rightarrow 2 \sin 2 x \cos x-3 \sin 2 x=2 \cos 2 x \cos x-3 \cos 2 x\)
\(\Rightarrow \sin 2 x(2 \cos x-3)=\cos 2 x(2 \cos x-3)\)
\(\Rightarrow \sin 2 x=\cos 2 x \quad[\because \cos x \neq 3 / 2]\)
\(\Rightarrow \tan 2 x=1 \Rightarrow 2 x=n \pi+\pi / 4\)
\(\Rightarrow x=\frac{n \pi}{2}+\frac{\pi}{8}\)

Hint

(c) \(\sqrt{3} \operatorname{cosec} 20^{\circ}-\sec 20^{\circ}\)
\(
\begin{aligned}
&=\frac{\sqrt{3}}{\sin 20^{\circ}}-\frac{1}{\cos 20^{\circ}}=\frac{\sqrt{3} \cos 20^{\circ}-\sin 20^{\circ}}{\sin 20^{\circ} \cos 20^{\circ}} \\
&=4\left[\frac{\left.\frac{\sqrt{3}}{2} \cos 20^{\circ}-\frac{1}{2} \sin 20^{\circ}\right]}{2 \sin 20^{\circ} \cos 20^{\circ}}\right] \\
&=4\left[\frac{\sin 60^{\circ} \cos 20^{\circ}-\cos 60^{\circ} \sin 20^{\circ}}{\sin \left(2 \times 20^{\circ}\right)}\right] \\
&=\frac{4 \sin 40^{\circ}}{\sin 40^{\circ}}=4
\end{aligned}
\)

Hint

(c) Given: \(\sin x+\cos x=1\)
\(
\Rightarrow \frac{1}{\sqrt{2}} \sin x+\frac{1}{\sqrt{2}} \cos =\frac{1}{\sqrt{2}}
\)
\(
\begin{aligned}
&\Rightarrow \sin x \cos \frac{\pi}{4}+\cos x \sin \frac{\pi}{4}=\sin \frac{\pi}{4} \\
&\Rightarrow \sin (x+\pi / 4)=\sin \pi / 4 \\
&\Rightarrow x+\pi / 4=n \pi+(-1)^{n} \pi / 4, n \in Z \text { (the set of integers) } \\
&\Rightarrow x=n+(-1)^{n} \pi / 4-\pi / 4
\end{aligned}
\)
where \(n=0, \pm 1, \pm 2, \ldots\)

Hint

(a) Given :
\(2 \cos ^{2}\left(\frac{x}{2}\right) \sin ^{2} x=x^{2}+\frac{1}{x^{2}}\) where \(0<x \leq \frac{\pi}{2}\)
\(
\begin{aligned}
\text { LHS }=& 2 \cos ^{2} \frac{x}{2} \sin ^{2} x=(1+\cos x) \sin ^{2} x \\
& \because 1+\cos x<2 \text { and } \sin ^{2} x \leq 1 \text { for } 0<x \leq \frac{\pi}{2} \\
\therefore(1+\cos x) \sin ^{2} x<2
\end{aligned}
\)
And R.H.S. \(=x^{2}+\frac{1}{x^{2}} \geq 2\)
Thus for \(0<x \leq \frac{\pi}{2}\), given equation is not possible

Hint

\(
\begin{aligned}
&\text { (b) } \frac{5}{4} \cos ^{2} 2 x+\cos ^{4} x+\sin ^{4} x+\cos ^{6} x+\sin ^{6} x=2 \\
&\Rightarrow \quad \frac{5}{4} \cos ^{2} 2 x+1-\frac{1}{2} \sin ^{2} 2 x+1-\frac{3}{4} \sin ^{2} 2 x=2 \\
&\Rightarrow \quad \frac{5}{4}\left(\cos ^{2} 2 x-\sin ^{2} 2 x\right)=0 \Rightarrow \cos 4 x=0 \\
&\Rightarrow \quad 4 x=(2 n+1) \frac{\pi}{2} \text { or } \quad x=(2 n+1) \frac{\pi}{8}
\end{aligned}
\)
For \(x \in[0,2 \pi]\), \(\mathrm{n}\) can take values 0 to 7 Hence, there are 8 solutions.

Hint

\(
\text { (c) } \frac{1}{\sin \frac{\pi}{n}}-\frac{1}{\sin \frac{3 \pi}{n}}=\frac{1}{\sin \frac{2 \pi}{n}}
\)
\(
\begin{aligned}
&\Rightarrow \frac{\sin \frac{3 \pi}{n}-\sin \frac{\pi}{n}}{\sin \frac{\pi}{n} \sin \frac{3 \pi}{n}}=\frac{1}{\sin \frac{2 \pi}{n}} \Rightarrow \frac{2 \cos \frac{2 \pi}{n} \sin \frac{\pi}{n}}{\sin \frac{\pi}{n} \sin \frac{3 \pi}{n}}=\frac{1}{\sin \frac{2 \pi}{n}} \\
&\Rightarrow \quad 2 \sin \frac{2 \pi}{n} \cos \frac{2 \pi}{n}=\sin \frac{3 \pi}{n} \Rightarrow \sin \frac{4 \pi}{n}-\sin \frac{3 \pi}{n}=0 \\
&\Rightarrow \quad 2 \cos \frac{7 \pi}{2 n} \sin \frac{\pi}{2 n}=0 \Rightarrow \cos \frac{7 \pi}{2 n}=0 \text { or } \sin \frac{\pi}{2 n}=0 \\
&\Rightarrow \quad \frac{7 \pi}{2 n}=(2 k+1) \frac{\pi}{2} \text { or } \frac{\pi}{2 n}=2 \mathrm{k} \pi, \text { where } k \in Z \\
&\Rightarrow \quad n=\frac{7}{2 k+1} \text { or } \mathrm{n}=\frac{1}{4 k}
\end{aligned}
\)
( \(n=\frac{1}{4 k}\) not possible for any integral value of \(k\) )
As \(n>3\); for \(k=0\), we get \(n=7\).

Hint

(d) From the figure,
\(
2 \cos \frac{\pi}{k}+2 \cos \frac{\pi}{2 k}=\sqrt{3}+1
\)
\(
\begin{gathered}
\Rightarrow 2 \times 2 \cos ^{2} \frac{\pi}{2 k}+2 \cos \frac{\pi}{2 k}-2 \\
=\sqrt{3}+1
\end{gathered}
\)
\(
\Rightarrow 4 \cos ^{2} \frac{\pi}{2 k}+2 \cos \frac{\pi}{2 k}-(3+\sqrt{3})=0
\)
\(
\begin{aligned}
&\Rightarrow \cos \frac{\pi}{2 k}=\frac{-2 \pm \sqrt{4+16(3+\sqrt{3})}}{8}=\frac{-1 \pm \sqrt{13+4 \sqrt{3}}}{4} \\
&=\frac{-1 \pm(2 \sqrt{3}+1)}{4}=\frac{\sqrt{3}}{2} \text { or }-\left(\frac{\sqrt{3}+1}{2}\right)
\end{aligned}
\)
As \(\frac{\pi}{2 k}\) is an acute angle, \(\cos \frac{\pi}{2 k}=\frac{\sqrt{3}}{2}=\cos \frac{\pi}{6} \Rightarrow k=3\)

Hint

(b) \(\tan \theta=\cot 5 \theta, \theta \neq \frac{n \pi}{5}\)
\(
\begin{aligned}
&\Rightarrow \cos \theta \cos 5 \theta-\sin 5 \theta \sin \theta=0 \Rightarrow \cos 6 \theta=0 \\
&\Rightarrow 6 \theta=\frac{-5 \pi}{2}, \frac{-3 \pi}{2}, \frac{-\pi}{2}, \frac{\pi}{2}, \frac{3 \pi}{2}, \frac{5 \pi}{2} \\
&\Rightarrow \theta=\frac{-5 \pi}{12}, \frac{-\pi}{4}, \frac{-\pi}{12}, \frac{\pi}{12}, \frac{\pi}{4}, \frac{5 \pi}{12}
\end{aligned}
\)
Again \(_{\sin 2 \theta} 2=\cos 4 \theta=1-2 \sin ^{2} 2 \theta\)
\(
\begin{aligned}
&\Rightarrow 2 \sin ^{2} 2 \theta+\sin 2 \theta-1=0 \Rightarrow \sin 2 \theta=-1, \frac{1}{2} \\
&\Rightarrow 2 \theta=\frac{-\pi}{2}, \frac{\pi}{6}, \frac{5 \pi}{6} \Rightarrow \theta=\frac{-\pi}{4}, \frac{\pi}{12}, \frac{5 \pi}{12}
\end{aligned}
\)
So, common solutions are \(\theta=\frac{-\pi}{4}, \frac{\pi}{12}\) and \(\frac{5 \pi}{12}\)
\(\therefore\) Number of solutions \(=3\).

Hint

(d) Given equations are
\(x y z \sin 3 \theta=(y+z) \cos 3 \theta\)…(i)
\(x y z \sin 3 \theta=2 z \cos 3 \theta+2 y \sin 3 \theta \quad\)…(ii)
\(x y z \sin 3 \theta=(y+2 z) \cos 3 \theta+y \sin 3 \theta\)…(iii)
On subtracting eq. (ii) from (i), we get
\((\cos 3 \theta-2 \sin \theta) y-(\cos 3 \theta) z=0\)…(iv)
On subtracting eq. (i) from (iii), we get \((\sin 3 \theta) y+(\cos 3 \theta) z=0\)…(v)
Eq. (iv) and (v) from homogeneous system of linear equation.
But \(y \neq 0, z \neq 0\)
\(
\begin{aligned}
&\therefore \frac{\cos 3 \theta-2 \sin 3 \theta}{\sin 3 \theta}=-\frac{\cos 3 \theta}{\cos 3 \theta} \Rightarrow \cos 3 \theta=\sin 3 \theta \\
&\Rightarrow \tan 3 \theta=1 \Rightarrow 3 \theta=n \pi+\frac{\pi}{4} \Rightarrow \theta=(4 n+1) \frac{\pi}{12}, n \in Z \\
&\text { For } \theta \in(0, \pi) \Rightarrow \theta=\frac{\pi}{12}, \frac{5 \pi}{12}, \frac{3 \pi}{4}
\end{aligned}
\)
Three such solutions are possible.

Hint

\(
\begin{aligned}
&\text { (c) } \log _{1 / 2}|\sin x|=2-\log _{1 / 2}|\cos x| \\
&\Rightarrow \quad \log _{1 / 2}|\sin x \cos x|=2 \\
&\Rightarrow \quad|\sin x \cos x|=\frac{1}{4} \\
&\Rightarrow \quad \sin 2 x=\pm \frac{1}{2}
\end{aligned}
\)

\(\text { Hence, total number of solutions }=8 \text {. }\)

Hint

(c) Given : \(\sqrt{3} a \cos x+2 b \sin x=c\)
which has two roots \(\alpha\) and \(\beta\), such that \(\alpha+\beta=\frac{\pi}{3}\)
\(\therefore \sqrt{3} \mathrm{a} \cos \alpha+2 \mathrm{~b} \sin \alpha=\mathrm{c}\)…(i)
and \(\sqrt{3}\) a \(\cos \beta+2 b \sin \beta=c\)…(ii)
On subtracting equation (ii) from (i),
\(
\begin{aligned}
& \sqrt{3} \text { a }(\cos \alpha-\cos \beta)+2 b(\sin \alpha-\sin \beta)=0 \Rightarrow \\
&-\sqrt{3} \mathrm{a} 2 \sin \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2}+2 b \cdot 2 \cos \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2}=0 \\
\Rightarrow &-2 \sqrt{3} \mathrm{a} \sin \frac{\pi}{6}+4 \mathrm{~b} \cos \frac{\pi}{6}=0 \\
\Rightarrow &-2 \sqrt{3} \mathrm{a} \times \frac{1}{2}+4 \mathrm{~b} \frac{\sqrt{3}}{2}=0 \Rightarrow \frac{\mathrm{b}}{\mathrm{a}}=\frac{1}{2}=0.5
\end{aligned}
\)

Hint

(a) \(\cos ^{7} x=1-\sin ^{4} x=\left(1-\sin ^{2} x\right)\left(1+\sin ^{2} x\right)=\cos ^{2} x\left(1+\sin ^{2} x\right)\)
\(\therefore \cos x=0 \Rightarrow x=\pi / 2,-\pi / 2\)
or \(\cos ^{5} x=1+\sin ^{2} x \Rightarrow \cos ^{5} x-\sin ^{2} x=1\)
Now maximum value of each \(\cos x\) or \(\sin x\) is 1.
Hence the above equation will hold when
\(\cos x=1\) and \(\sin x=0\)
Both these imply \(x=0\)
\(
\therefore x=-\frac{\pi}{2}, \frac{\pi}{2}, 0
\)

Hint

(b)

\(
\begin{aligned}
&\tan ^{2} \theta+\sec 2 \theta=1 \\
&\left.t^{2}+\frac{1+t^{2}}{1-t^{2}}=1, \text { [but } t=\tan \theta\right] \\
&\Rightarrow \quad t^{2}\left(t^{2}-3\right)=0 \Rightarrow \tan \theta=0, \pm \sqrt{3} \\
&\Rightarrow \quad \theta=n \pi \text { and } \theta=n \pi \pm \pi / 3
\end{aligned}
\)

Hint

(a) We know that A.M. \(\geq\) G.M.
\(\Rightarrow\) Minimum value of \(\mathrm{AM}\). is obtained when \(\mathrm{AM}=\mathrm{GM}\)
\(\Rightarrow\) The quantities whose \(\mathrm{AM}\) is being taken are equal
i.e., \(\cos \left(\alpha+\frac{\pi}{2}\right)=\cos \left(\beta+\frac{\pi}{2}\right)\)
\(
=\cos \left(\gamma+\frac{\pi}{2}\right)
\)
\(\Rightarrow \sin \alpha=\sin \beta=\sin \gamma\)
Also \(\alpha+\beta+\gamma=360^{\circ} \Rightarrow \alpha=\beta=\gamma=120^{\circ}=2 \pi / 3\)
\(\therefore\) Min value of A.M.
\(
\begin{aligned}
&=\frac{\cos \left(\frac{2 \pi}{3}+\frac{\pi}{2}\right)+\cos \left(\frac{2 \pi}{3}+\frac{\pi}{2}\right)+\cos \left(\cos \left(\frac{2 \pi}{3}+\frac{\pi}{2}\right)\right)}{3} \\
&=\frac{-3 \sin \frac{2 \pi}{3}}{3}=-\frac{\sqrt{3}}{2}
\end{aligned}
\)

Hint

(c) Given: \(2 \sin ^{2} x-3 \sin x+1 \geq 0\)
\(\Rightarrow(2 \sin x-1)(\sin x-1) \geq 0\)
\(\Rightarrow\left(\sin x-\frac{1}{2}\right)(\sin x-1) \geq 0 \Rightarrow \sin x \leq \frac{1}{2}\) or \(\sin x \geq 1\)
But we know that \(0 \leq \sin x \leq 1\) for \(x \in[0, \pi]\)
\(
\begin{aligned}
&\Rightarrow \quad 0 \leq \sin x \leq \frac{1}{2} \text { or } \sin x=1 \\
&\Rightarrow \quad x \in[0, \pi / 6] \cup[5 \pi / 6, \pi] \text { or } x=\pi / 2
\end{aligned}
\)
On combining, we get \(x \in[0, \pi / 6] \cup\{\pi / 2\} \cup[5 \pi / 6, \pi]\)

Hint

(a) Given equations: \(x+y=2 \pi / 3\)
and \(\cos x+\cos y=3 / 2\)
From eq. (ii), \(2 \cos \frac{x+y}{2} \cos \frac{x-y}{2}=\frac{3}{2}\)
\(
\begin{aligned}
&\Rightarrow \quad 2 \cos \frac{\pi}{3} \cos \frac{x-y}{2}=\frac{3}{2} \quad \text { [From eq. (i)] } \\
&\Rightarrow \quad 2 . \frac{1}{2} \cos \frac{x-y}{2}=\frac{3}{2} \Rightarrow \cos \frac{x-y}{2}=\frac{3}{2}
\end{aligned}
\)
which is not possible because \(-1 \leq \cos \theta \leq 1\).
Hence, solution of given equations is \(\phi\).

Hint

(b) (False) Given : \(\sin ^{4} \theta-2 \sin ^{2} \theta-1=0\)
\(
\therefore \quad D=4+4=8
\)
\(
\because \sin ^{2} \theta=\frac{2 \pm \sqrt{8}}{2}=1 \pm \sqrt{2} \text {. }
\)
\(\because\) value of \(\sin ^{2} \theta\) is +ve, \(\quad \therefore \sin ^{2} \theta=\sqrt{2}+1\)
\(\because-1 \leq \sin \theta \leq 1, \quad \therefore \sin ^{2} \theta \neq \sqrt{2}+1\)
Hence, there is no value of \(\theta\), which satisfy the given equation. Hence, the statement is false.

Hint

(a, c) If we consider \(\tan \alpha / 2=x\) and \(\tan \beta / 2=y\), then \(2(\cos \beta-\cos \alpha)+\cos \alpha \cos \beta=1\)
\(
\begin{aligned}
&\Rightarrow \quad 2\left[\frac{1-y^{2}}{1+y^{2}}-\frac{1-x^{2}}{1+x^{2}}\right]=1-\frac{\left(1-x^{2}\right)\left(1-y^{2}\right)}{\left(1+x^{2}\right)\left(1+y^{2}\right)} \\
&\Rightarrow \quad 2\left[\left(1+x^{2}\right)\left(1-y^{2}\right)-\left(1-x^{2}\right)\left(1+y^{2}\right)\right] \\
&\Rightarrow \quad 4\left(x^{2}-y^{2}\right)=2\left(x^{2}+y^{2}\right) \\
&\Rightarrow \quad x^{2}=3 y^{2} \Rightarrow x=\pm \sqrt{3} y \\
&\Rightarrow \quad \tan \frac{\alpha}{2} \pm \sqrt{3} \tan \frac{\beta}{2}=0
\end{aligned}
\)

Hint

(c) Let \(f(x)=x^{2}-x \sin x-\cos x\)
\(\therefore f(x)=2 x-x \cos x=x(2-\cos x)\)
\(\therefore f\) is increasing on \((0, \infty)\) and decreasing on \((-\infty, 0)\)
Also \(\lim _{x \rightarrow \infty} f(x)=\infty, \lim _{x \rightarrow-\infty} f(x)=\infty\) and \(f(0)=-1\)
\(\therefore \quad y=f(x)\) meets \(x\)-axis twice.
i.e., \(f(x)=0\) has two points in \((-\infty, \infty)\).

Hint

(c,d) Given:
\(
\begin{aligned}
&\frac{1}{\sin (\pi / 4)}\left[\frac{\sin (\theta+\pi / 4-\theta)}{\sin \theta \sin (\theta+\pi / 4)}+\frac{\sin (\theta+\pi / 2-(\theta+\pi / 4))}{\sin (\theta+\pi / 4) \cdot \sin (\theta+\pi / 2)}+\ldots+\right. \\
&\left.\frac{\sin ((\theta+3 \pi / 2)-(\theta+5 \pi / 4))}{\sin (\theta+3 \pi / 2) \cdot \sin (\theta+5 \pi / 4)}\right]=4 \sqrt{2}
\end{aligned}
\)
Using Formula for splitting numerator of first term:
\(
\sin (\theta+\pi / 4-\theta)=\sin (\theta+\pi / 4) \cos (\theta)-\cos (\theta+\pi / 4) \sin (\theta)
\)
\(
\begin{aligned}
&\Rightarrow \sqrt{2}[\cot \theta-\cot (\theta+\pi / 4)+\cot (\theta+\pi / 4)-\cot (\theta+\pi / 2)+\ldots+\cot (\theta+5 \pi / 4)-\cot (\theta+ \\
&3 \pi / 2)]=4 \sqrt{2}
\end{aligned}
\)
\(
\Rightarrow \tan \theta+\cot \theta=4 \Rightarrow \tan \theta=2 \pm \sqrt{3}
\)
\(
\Rightarrow \theta=\frac{\pi}{12} \text { or } \frac{5 \pi}{12}
\)

Hint

(c) \(3 \sin ^{2} x-7 \sin x+2=0\), put \(\sin x=s\)
\(\Rightarrow(s-2)(3 s-1)=0\)
\(\because \quad s=2\) is not possible, \(\quad \therefore s=1 / 3 \Rightarrow \sin x=\frac{1}{3}\)
Number of solutions of \(\sin x=\frac{1}{3}\) from the following graph is 6 between \([0,5 \pi]\)

Hint

\(
\text { (a, c) }\left|\begin{array}{rrr}
1+\sin ^{2} \theta & \cos ^{2} \theta & 4 \sin 4 \theta \\
\sin ^{2} \theta & 1+\cos ^{2} \theta & 4 \sin 4 \theta \\
\sin ^{2} \theta & \cos ^{2} \theta & 1+4 \sin 4 \theta
\end{array}\right|=0
\)
Applying \(C_{1} \rightarrow C_{1}+C_{2}\)
\(
\left|\begin{array}{rrr}
2 & \cos ^{2} \theta & 4 \sin 4 \theta \\
2 & 1+\cos ^{2} \theta & 4 \sin 4 \theta \\
1 & \cos ^{2} \theta & 1+4 \sin 4 \theta
\end{array}\right|=0
\)
Applying \(R_{1} \rightarrow R_{1}-R_{2} ; R_{2} \rightarrow R_{2}-R_{3}\)
\(
\left|\begin{array}{rrr}
0 & -1 & 0 \\
1 & 1 & -1 \\
1 & \cos ^{2} \theta & 1+4 \sin 4 \theta
\end{array}\right|=0
\)
\(
\begin{aligned}
&\Rightarrow \quad[1+4 \sin 4 \theta+1]=0 \\
&\Rightarrow \quad 2(1+2 \sin 4 \theta)=0 \Rightarrow \sin 4 \theta=-\frac{1}{2} \\
&\Rightarrow 4 \theta=\pi+\pi / 6 \text { or } 2 \pi-\pi / 6 \\
&\Rightarrow 4 \theta=7 \pi / 6 \text { or } 11 \pi / 6 \\
&\Rightarrow \quad \theta=7 \pi / 24 \text { or } 11 \pi / 24
\end{aligned}
\)

Hint

(d) \(a_{1}+a_{2} \cos 2 x+a_{3} \sin ^{2} x=0, \forall x\)
For \(x=0\) and \(x=\pi / 2\), we get \(a_{1}+a_{2}=0\)….(i)
and \(a_{1}-a_{2}+a_{3}=0\)
\(\Rightarrow a_{2}=-a_{1}\) and \(a_{3}=-2 a_{1}\)
\(\therefore\) The given equation becomes
\(
\begin{aligned}
&a_{1}-a_{1} \cos 2 x-2 a_{1} \sin ^{2} x=0, \quad \forall x \\
&\Rightarrow a_{1}\left(1-\cos 2 x-2 \sin ^{2} x\right)=0, \forall x \\
&\Rightarrow a_{1}\left(2 \sin ^{2} x-2 \sin ^{2} x\right)=0, \forall x
\end{aligned}
\)
The above is satisfied for all values of \(a_{1}\).
Hence, infinite number of triplets \(\left(a_{1},-a_{1},-2 a_{1}\right)\) are possible.

Hint

\(
\begin{aligned}
&f(x)=\sin (\pi \cos x) \\
& X=\left\{x: f^{\prime}(x)=0\right\} \\
&f(x)=0 \Rightarrow \sin (\pi \cos x)=0 \Rightarrow \cos x=n \Rightarrow \cos x=1,-1,0 \Rightarrow x=\frac{n \pi}{2} \\
&X=\left\{\frac{n \pi}{2}: n \in N\right\}=\left\{\frac{\pi}{2}, \pi, \frac{3 \pi}{2}, 2 \pi \ldots .\right\} \\
&g(x)=\cos (2 \pi \sin x) \\
&Z=\{x: g(x)=0\} \\
&\cos (2 \pi \sin x)=0 \Rightarrow 2 \pi \sin x=(2 n+1) \frac{\pi}{2} \Rightarrow \sin x=\frac{(2 n+1)}{4} \\
&\sin x=-\frac{1}{4}, \frac{1}{4}, \frac{-3}{4}, \frac{3}{4} \\
&Z=\left\{n \pi \pm \sin -1\left(\frac{1}{4}\right), n \pi \pm \sin ^{-1}\left(\frac{3}{4}\right), n \in I\right\} \\
&Y=\{x: f(x)=0\} \\
&f(x)=\sin (\pi \cos x) \Rightarrow f^{\prime}(x)=\cos (\pi \cos x) \cdot(-\pi \sin x)=0 \\
&\sin x=0 \Rightarrow x=n \pi . \\
&\cos (\pi \cos x)=0 \pi \cos x=(2 n+1) \frac{\pi}{2} \Rightarrow \cos =\frac{(2 n+1)}{2} \Rightarrow \cos x=-\frac{1}{2}, \frac{1}{2} \\
&Y=\left\{n \pi, n \pi \pm \frac{\pi}{3}\right\}=\left\{\frac{\pi}{3}, \frac{2 \pi}{3}, \pi, \frac{4 \pi}{3}, \frac{5 \pi}{3}, 2 \pi, \ldots \ldots\right\} \\
&W=\left\{x: g^{\prime}(x)=0\right\}
\end{aligned}
\)

\(
\begin{aligned}
&g(x)=\cos (2 \pi \sin x) \Rightarrow g^{\prime}(x)=-\sin (2 \pi \sin x) \cdot(2 \pi \cos x)=0 \\
&\cos x=0 \Rightarrow x=(2 n+1) \frac{\pi}{2} \\
&\sin (2 \pi \sin x)=0 \Rightarrow 2 \pi \sin x=n \pi \Rightarrow \sin x=\frac{n}{2}=-1,-\frac{1}{2}, 0, \frac{1}{2}, 1 \\
&W=\left\{\frac{n \pi}{2}, n \pi \pm \frac{\pi}{6}, n \in 1\right\}=\left\{\frac{\pi}{6}, \frac{\pi}{2}, \frac{5 \pi}{6}, \pi, \frac{7 \pi}{6}, \frac{3 \pi}{2}, \ldots . .\right\}
\end{aligned}
\)

Hint

(d)

\(
\begin{aligned}
&f^{\prime}(\mathrm{x})=0 \quad \Rightarrow \quad \cos (\pi \cos x)(-\pi \sin x)=0 \\
&\quad \Rightarrow \quad \cos (\pi \cos x)=0, \sin x=0 \\
&\Rightarrow \quad \pi \cos x=(2 n-1) \frac{\pi}{2}, x=n \pi \\
&\Rightarrow \quad \cos x=(2 n-1) \frac{1}{2}, x=\pi, 2 \pi, 3 \pi, \ldots \ldots \\
&\Rightarrow \quad \cos x=\frac{-1}{2}, \frac{1}{2} . \\
&\Rightarrow \quad x=\frac{\pi}{3}, \frac{2 \pi}{3}, \frac{4 \pi}{3}, \frac{5 \pi}{3}, \frac{7 \pi}{3}, \frac{8 \pi}{3}, \frac{10 \pi}{3}, \frac{11 \pi}{3}, \frac{13 \pi}{3}, \ldots . . \\
&\therefore Y=\left\{\frac{\pi}{3}, \frac{2 \pi}{3}, \pi, \frac{4 \pi}{3}, \frac{5 \pi}{3}, 2 \pi, \ldots \ldots \ldots\right\} . \\
&\therefore \quad(\mathrm{II})-\mathrm{Q}, \mathrm{T} .
\end{aligned}
\)

Hint

Given : \((1-\tan \theta)(1+\tan \theta) \sec ^{2} \theta+2^{\tan ^{2} \theta}=0\) \(\Rightarrow \quad\left(1-\tan ^{2} \theta\right)\left(1+\tan ^{2} \theta\right)+2^{\tan ^{2} \theta}=0\)
Put \(\tan ^{2} \theta=t\)
\(\therefore \quad(1-t)(1+t)+2^{t}=0\) or \(1-t^{2}+2^{t}=0\)
It is clearly satisfied by \(t=3\).
As \(-8+8=0 \quad \therefore \quad \tan ^{2} \theta=3\)
\(\therefore \quad \theta=\pm \pi / 3\) in the given interval.

Hint

\(
\begin{aligned}
&8^{\left(1+|\cos x|+\left|\cos ^{2} x\right|+\left|\cos ^{3} x\right|+\ldots .\right)}=4^{3} \\
&\Rightarrow 2^{3\left(1+|\cos x|+\left|\cos ^{2} x\right|+\left|\cos ^{3} x\right|+\ldots\right)}=2^{6} \\
&\Rightarrow 3\left(1+|\cos x|+\left|\cos ^{2} x\right|+\left|\cos ^{3} x\right|+\ldots\right)=6 \\
&\Rightarrow 1+|\cos x|+\left|\cos ^{2} x\right|+\left|\cos ^{3} x\right|+\ldots .=2 \\
&\Rightarrow \frac{1}{1-|\cos x|}=2 \\
&\Rightarrow 1-\cos x=1 / 2 \Rightarrow|\cos x|=\frac{1}{2} \\
&\Rightarrow x=\pi / 3,-\pi / 3,2 \pi / 3,-2 \pi / 3, \ldots .
\end{aligned}
\)
The values of \(x \in(-\pi, \pi)\) are \(\pm \pi / 3, \pm 2 \pi / 3\).

Hint

Given : \(4 \cos ^{2} x \sin x-2 \sin ^{2} x=3 \sin x\)
\(\Rightarrow 4 \cos ^{2} x \sin x-2 \sin ^{2} x-3 \sin x=0\)
\(\Rightarrow 4\left(1-\sin ^{2} x\right) \sin x-2 \sin ^{2} x-3 \sin x=0\)
\(\Rightarrow \sin x\left[4 \sin ^{2} x+2 \sin x-1\right]=0\)
\(\therefore \quad\) Either \(\sin x=0\) or \(4 \sin ^{2} x+2 \sin x-1=0\)
If \(\sin x=0\), then \(x=n \pi\)
\(\Rightarrow\) If \(4 \sin ^{2} x+2 \sin x-1=0\), then \(\sin x=\frac{-1 \pm \sqrt{5}}{4}\)
If \(\sin x=\frac{-1 \pm \sqrt{5}}{4}=\sin 18^{\circ}=\sin \frac{\pi}{10}\), then
\(x=n x+(-1)^{4} \frac{\pi}{10}\)
If \(\sin x=-\left(\frac{\sqrt{5}+1}{4}\right)=\sin \left(-54^{\circ}\right)=\sin \left(\frac{-3 \pi}{10}\right)\),
then \(x=n \pi+(-1)^{n}\left(\frac{-3 \pi}{10}\right)\)
Hence, \(x=n \pi, n \pi+(-1)^{n} \frac{\pi}{10}\) or \(n \pi+(-1)^{n}\left(\frac{-3 \pi}{10}\right)\),
where \(n\) is some integer

Hint

(a) Given: \(y=\frac{1}{\sqrt{2}}(\sin x+\cos x)=\sin \left(x+\frac{\pi}{4}\right)\)
To draw the graph of \(y=\sin \left(x+\frac{\pi}{4}\right)\), we first draw the graph of \(y=\sin x\) and then on shifting it by \(-\frac{\pi}{4}\).

\(
\begin{aligned}
&\text { (b) } \cos (\alpha+\beta)=\frac{4}{5} \\
&\Rightarrow \tan (\alpha+\beta)=\frac{3}{4}, 0<\alpha, \beta<\frac{\pi}{4} \\
&\sin (\alpha-\beta)=\frac{5}{13} \Rightarrow \tan (\alpha-\beta)=\frac{5}{12} \\
&=\frac{\tan (\alpha+\beta)+\tan (\alpha-\beta)}{1-\tan (\alpha+\beta) \tan (\alpha-\beta)} =\tan 2 \alpha=\tan [(\alpha+\beta)+(\alpha-\beta)] \\
&=\frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{3}{4} \times \frac{5}{12}}=\frac{9+5}{12} \times \frac{16}{11}=\frac{56}{33}
\end{aligned}
\)

Hint

(d) We start by recognizing that \(\sin ^2 x+\cos ^2 x=1\). Substituting \(\sin ^2 x=1-\cos ^2 x\) into the original equation gives:
\(
4\left(1-\cos ^2 x\right)-4 \cos ^3 x+9-4 \cos x=0
\)
Rearranging and simplifying this equation, we have:
\(
\begin{aligned}
& 4-4 \cos ^2 x-4 \cos ^3 x+9-4 \cos x=0 \\
& 4 \cos ^3 x+4 \cos ^2 x+4 \cos x-13=0 \\
& 4 \cos ^3 x+4 \cos ^2 x+4 \cos x=13
\end{aligned}
\)
Observing the bounds given, \(x \in[-2 \pi, 2 \pi]\), we want to find how many solutions satisfy this cubic equation in terms of \(\cos x\). However, we note that the left-hand side (LHS) of the equation, representing a combination of cosines, could at most approach a maximum sum when \(\cos x=1\), that being \(4(1)+4(1)+4(1)=12\). Yet, we have the equation set to equal 13 , which is impossible given the maximum sum of the LHS can only be 12.
The above reasoning indicates that, within the domain specified, there is no value of \(x\) for which the equation holds true. Therefore, the number of solutions to the equation is zero.

Hint

(c)

\(
\begin{aligned}
& \cos 2 x+a \cdot \sin x=2 a-7 \\
& a(\sin x-2)=2(\sin x-2)(\sin x+2) \\
& \sin x=2, a=2(\sin x+2) \\
& \Rightarrow a \in[2,6] \\
& p=2 \quad q=6 \\
& r=\tan 9^{\circ}+\cot 9^{\circ}-\tan 27^{\circ}-\cot 27^{\circ} \\
& r=\frac{1}{\sin 9 \cdot \cos 9}-\frac{1}{\sin 27 \cdot \cos 27} \\
& =2\left[\frac{4}{\sqrt{5}-1}-\frac{4}{\sqrt{5}+1}\right] \\
& r=4 \\
& p \cdot q \cdot r=2 \times 6 \times 4=48
\end{aligned}
\)

Hint

(d)

\(
\begin{aligned}
& 2 \tan ^2 \theta-5 \sec \theta-1=0 \\
& \Rightarrow 2 \sec ^2 \theta-5 \sec \theta-3=0 \\
& \Rightarrow(2 \sec \theta+1)(\sec \theta-3)=0 \\
& \Rightarrow \sec \theta=-\frac{1}{2}, 3 \\
& \Rightarrow \cos \theta=-2, \frac{1}{3} \\
& \Rightarrow \cos \theta=\frac{1}{3}
\end{aligned}
\)
For 7 solutions \(n=13\)
\(
\begin{aligned}
& \text { So, } \sum_{ k =1}^{13} \frac{ k }{2^{ k }}= S \text { (say) } \\
& S =\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\ldots+\frac{13}{2^{13}} \\
& \frac{1}{2} S =\frac{1}{2^2}+\frac{1}{2^3}+\ldots+\frac{12}{2^{13}}+\frac{13}{2^{14}} \\
& \Rightarrow \frac{ S }{2}=\frac{1}{2} \cdot \frac{1-\frac{1}{2^{13}}}{1-\frac{1}{2}}-\frac{13}{2^{14}} \Rightarrow S =2 \cdot\left(\frac{2^{13}-1}{2^{13}}\right)-\frac{13}{2^{13}}=\frac{1}{2^{13}}\left(2^{14}-15\right)
\end{aligned}
\)

Hint

(c)

\(
4+5 \tan \theta=\sec \theta
\)
Squaring: \(24 \tan ^2 \theta+40 \tan \theta+15=0\)
\(
\tan \theta=\frac{-10 \pm \sqrt{10}}{12}
\)
and \(\tan \theta=-\left(\frac{10+\sqrt{10}}{12}\right)\) is Rejected.

Hint

(c)

\(
\begin{aligned}
& \frac{3 \cos 2 x+\cos ^3 2 x}{\cos ^6 x-\sin ^6 x}=x^3-x^2+6 \\
& \Rightarrow \frac{\cos 2 x\left(3+\cos ^2 2 x\right)}{\cos 2 x\left(1-\sin ^2 x \cos ^2 x\right)}=x^3-x^2+6 \\
& \Rightarrow \frac{4\left(3+\cos ^2 2 x\right)}{\left(4-\sin ^2 2 x\right)}=x^3-x^2+6 \\
& \Rightarrow \frac{4\left(3+\cos ^2 2 x\right)}{\left(3+\cos ^2 2 x\right)}=x^3-x^2+6 \\
& x^3-x^2+2=0 \Rightarrow(x+1)\left(x^2-2 x+2\right)=0
\end{aligned}
\)
so, sum of real solutions \(=-1\)

Hint

(b)

\(
\begin{aligned}
& 2 \sin ^3 x+2 \sin x \cdot \cos ^2 x+4 \sin x-4=0 \\
& 2 \sin ^3 x+2 \sin x \cdot\left(1-\sin ^2 x\right)+4 \sin x-4=0 \\
& 6 \sin x-4=0 \\
& \sin x=\frac{2}{3}
\end{aligned}
\)
\(n = 5\) (in the given interval)
\(
\begin{aligned}
& x^2+5 x+2=0 \\
& x=\frac{-5 \pm \sqrt{17}}{2}
\end{aligned}
\)
Required interval \((-\infty, 0)\)

Hint

(a)

\(
\begin{aligned}
& \text { Finding } \tan (A+B) \text { we get } \\
& \Rightarrow \tan ( A + B )= \\
& \frac{\tan A+\tan B}{1-\tan A \tan B}=\frac{\frac{1}{\sqrt{x\left(x^2+x+1\right)}}+\frac{\sqrt{x}}{\sqrt{x^2+x+1}}}{1-\frac{1}{x^2+x+1}} \\
& \Rightarrow \tan ( A + B )=\frac{(1+x)\left(\sqrt{x^2+x+1}\right)}{\left(x^2+x\right)(\sqrt{x})} \\
& \frac{(1+x)\left(\sqrt{x^2+x+1}\right)}{\left(x^2+x\right)(\sqrt{x})} \\
& \tan (A+B)=\frac{\sqrt{x^2+x+1}}{x \sqrt{x}}=\tan C \\
& A+B=C
\end{aligned}
\)

Hint

(b)

\(
\begin{aligned}
& 3 \sin \alpha \cos \beta+3 \sin \beta \cos \alpha \\
& =2 \sin \alpha \cos \beta-2 \sin \beta \cos \alpha \\
& 5 \sin \beta \cos \alpha=-\sin \alpha \cos \beta \\
& \tan \beta=-\frac{1}{5} \tan \alpha \\
& \tan \alpha=-5 \tan \beta
\end{aligned}
\)

Hint

(d)

\(
\begin{aligned}
& \cos 2 \theta \cdot \cos \frac{\theta}{2}=\cos 3 \theta \cdot \cos \frac{9 \theta}{2} \\
& \Rightarrow 2 \cos 2 \theta \cdot \cos \frac{\theta}{2}=2 \cos \frac{9 \theta}{2} \cdot \cos 3 \theta \\
& \Rightarrow \cos \frac{5 \theta}{2}+\cos \frac{3 \theta}{2}=\cos \frac{15 \theta}{2}+\cos \frac{3 \theta}{2} \\
& \Rightarrow \cos \frac{15 \theta}{2}=\cos \frac{5 \theta}{2} \\
& \Rightarrow \frac{15 \theta}{2}=2 k \pi \pm \frac{5 \theta}{2} \\
& 5 \theta=2 k \pi \text { or } 10 \theta=2 k \pi \\
& \theta=\frac{2 k \pi}{5} \theta=\frac{k \pi}{5} \\
& \therefore \theta=\left\{-\pi, \frac{-4 \pi}{5}, \frac{-3 \pi}{5}, \frac{-2 \pi}{5}, \frac{-\pi}{5}, 0, \frac{\pi}{5}, \frac{2 \pi}{5}, \frac{3 \pi}{5}, \frac{4 \pi}{5}, \pi\right\} \\
& m=5, n=5 \\
& \therefore m \cdot n=25
\end{aligned}
\)

Hint

(b)

\(
f(\theta)=3\left(\sin ^4\left(\frac{3 \pi}{2}-\theta\right)+\sin ^4(3 x+\theta)\right)-2\left(1-\sin ^2 2 \theta\right)
\)
\(
\begin{aligned}
& S=\left\{\theta \in[0, \pi]: f^{\prime}(\theta)=-\frac{\sqrt{3}}{2}\right\} \\
& \Rightarrow f(\theta)=3\left(\cos ^4 \theta+\sin ^4 \theta\right)-2 \cos ^2 2 \theta
\end{aligned}
\)
\(
\begin{aligned}
&\Rightarrow f(\theta)=3\left(1-\frac{1}{2} \sin ^2 2 \theta\right)-2 \cos ^2 2 \theta\\
&\begin{aligned}
\Rightarrow f(\theta) & =3-\frac{3}{2} \sin ^2 2 \theta-2 \cos ^2 \theta \\
& =\frac{3}{2}-\frac{1}{2} \cos ^2 2 \theta=\frac{3}{2}-\frac{1}{2}\left(\frac{1+\cos 4 \theta}{2}\right)
\end{aligned}
\end{aligned}
\)
\(
\begin{aligned}
& f(\theta)=\frac{5}{4}-\frac{\cos 4 \theta}{4} \\
& f^{\prime}(\theta)=\sin 4 \theta \\
& \Rightarrow f^{\prime}(\theta)=\sin 4 \theta=-\frac{\sqrt{3}}{2} \\
& \Rightarrow 4 \theta= n \pi+(-1)^{ n } \frac{\pi}{3} \\
& \Rightarrow \theta=\frac{ n \pi}{4}+(-1)^{ n } \frac{\pi}{12} \\
& \Rightarrow \theta=\frac{\pi}{12},\left(\frac{\pi}{4}-\frac{\pi}{12}\right),\left(\frac{\pi}{2}+\frac{\pi}{12}\right),\left(\frac{3 \pi}{4}-\frac{\pi}{12}\right) \\
& \Rightarrow 4 \beta=\frac{\pi}{4}+\frac{\pi}{2}+\frac{3 \pi}{4}=\frac{3 \pi}{2} \\
& \Rightarrow \beta=\frac{3 \pi}{8} \Rightarrow f(\beta)=\frac{5}{4}-\frac{\cos \frac{3 \pi}{2}}{4}=\frac{5}{4} \\
&
\end{aligned}
\)

Hint

(b)

\(
\begin{aligned}
& \tan (\pi \cos \theta)+\tan (\pi \sin \theta)=0 \\
& \tan (\pi \cos \theta)=-\tan (\pi \sin \theta) \\
& \tan (\pi \cos \theta)=\tan (-\pi \sin \theta) \\
& \pi \cos \theta=n \pi-\pi \sin \theta \\
& \sin \theta+\cos \theta=n \text { where } n \in I
\end{aligned}
\)
possible values are \(n =0,1\) and -1 because
\(
-\sqrt{2} \leq \sin \theta+\cos \theta \leq \sqrt{2}
\)
Now it gives \(\theta \in\left\{0, \frac{\pi}{2}, \frac{3 \pi}{4}, \frac{7 \pi}{4}, \frac{3 \pi}{2}, \pi\right\}\)
\(
\text { So } \sum_{\theta=S} \sin ^2\left(\theta+\frac{\pi}{4}\right)=2(0)+4\left(\frac{1}{2}\right)=2
\)

Hint

(d)

\(
\lambda=\cos ^2 2 x-2 \sin ^4 x-2 \cos ^2 x
\)
convert all in to \(\cos x\).
\(
\begin{aligned}
\lambda & =\left(2 \cos ^2 x-1\right)^2-2\left(1-\cos ^2 x\right)^2-2 \cos ^2 x \\
& =4 \cos ^4 x-4 \cos ^2 x+1-2\left(1-2 \cos ^2 x+\cos ^4 x\right)- \\
& 2 \cos ^2 x \\
& =2 \cos ^4 x-2 \cos ^2 x+1-2 \\
& =2 \cos ^4 x-2 \cos ^2 x-1 \\
& =2\left[\cos ^4 x-\cos ^2 x-\frac{1}{2}\right] \\
& =2\left[\left(\cos ^2 x-\frac{1}{2}\right)^2-\frac{3}{4}\right] \\
\lambda_{\max } & =2\left[\frac{1}{4}-\frac{3}{4}\right]=2 \times\left(-\frac{2}{4}\right)=-1 \text { (max Value) } \\
\lambda_{\min } & =2\left[0-\frac{3}{4}\right]=-\frac{3}{2}(\text { MinimumValue) }
\end{aligned}
\)
So, Range \(=\left[-\frac{3}{2},-1\right]\)

Hint

(a)

\(
\begin{aligned}
& \tan 15^{\circ}=2-\sqrt{3} \\
& \frac{1}{\tan 75^{\circ}}=\cot 75^{\circ}=2-\sqrt{3} \\
& \frac{1}{\tan 105^{\circ}}=\cot \left(105^{\circ}\right)=-\cot 75^{\circ}=\sqrt{3}-2 \\
& \tan 195^{\circ}=\tan 15^{\circ}=2-\sqrt{3} \\
\therefore \quad & 2(2-\sqrt{3})=2 a \Rightarrow a =2-\sqrt{3} \\
\Rightarrow & a +\frac{1}{ a }=4
\end{aligned}
\)

Hint

(d)

\(
\begin{aligned}
& \log _{\cos x} \cot x+4 \log _{\sin x} \tan x=1 \\
& \Rightarrow \frac{\ln \cos x-\ln \sin x}{\ln \cos x}+4 \frac{\ln \sin x-\ln \cos x}{\ln \sin x}=1 \\
& \Rightarrow(\ln \sin x)^2-4(\ln \sin x)(\ln \cos x)+4(\ln \cos x)^2= \\
& \Rightarrow \ln \sin x=2 \ln \cos x \\
& \Rightarrow \sin ^2 x+\sin x-1=0 \Rightarrow \sin x=\frac{-1+\sqrt{5}}{2} \\
& \therefore \alpha+\beta=4
\end{aligned}
\)