Trigonometric Ratios & Identities

$ \begin{array}{r} \text { }\sin \frac{\pi}{12} \cdot \sin \frac{2 \pi}{12} \cdot \sin \frac{3 \pi}{12} \cdot \sin \frac{4 \pi}{12} \cdot \sin \frac{5 \pi}{12} \cdot \sin \frac{6 \pi}{12} \\ =\sin 15^{\circ} \cdot \sin 30^{\circ} \cdot \sin 45^{\circ} \cdot \sin 60^{\circ} \sin 75^{\circ} \cdot \sin 90^{\circ} \end{array} $

$ \begin{aligned} & =\frac{\sqrt{6}-\sqrt{2}}{4} \times \frac{1}{2} \times \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} \times \frac{\sqrt{6}+\sqrt{2}}{4} \times 1 \\ & =\frac{(\sqrt{6})^2-(\sqrt{2})^2}{4 \times 4} \times \frac{\sqrt{3}}{4 \sqrt{2}} \\ & =\frac{1}{4} \times \frac{\sqrt{3}}{4 \sqrt{2}}=\frac{\sqrt{3}}{16 \sqrt{2}} \end{aligned} $

$ \begin{aligned} & \text { }(\sin A+\sin D)+(\sin B+\sin C) \\ & \qquad \begin{aligned} =2 \sin \left(\frac{A+D}{2}\right) & \cos \left(\frac{A-D}{2}\right) + 2 \sin \left(\frac{B+C}{2}\right) \cos \left(\frac{B-C}{2}\right) \end{aligned} \end{aligned} $

$ \begin{aligned} & \because A+B+C+D=2 \pi \\ & \Rightarrow B+C=2 \pi-(A+D) \\ & \begin{aligned} = & 2 \sin \left(\frac{A+D}{2}\right) \cos \left(\frac{A-D}{2}\right) +2 \sin \left(\pi-\frac{A+D}{2}\right) \cos \left(\frac{B-C}{2}\right) \end{aligned} \end{aligned} $

$ \begin{array}{r} =2 \sin \left(\frac{A+D}{2}\right)\left[\cos \left(\frac{A-D}{2}\right)+\cos \left(\frac{B-C}{2}\right)\right.] \\ =2 \sin \left(\frac{A+D}{2}\right)\left[2 \cos \left(\frac{\left(\frac{A-D+B-C}{2}\right)}{2}\right)\right. \left.\cos \left(\frac{\left(\frac{A-D-B+C}{2}\right)}{2}\right)\right] \end{array} $

$ \begin{array}{r} =4 \sin \left(\frac{A+D}{2}\right)\left[\cos \left(\frac{A+B-(C+D)}{4}\right)\right. \left.\cos \left(\frac{A+C-(B+D)}{4}\right)\right] \end{array} $

$ \begin{aligned} & \because \quad C+D=2 \pi-(A+B) \\ & \text { and } B+D=2 \pi-(A+C) \\ & =4 \sin \left(\frac{A+D}{2}\right)\left[\cos \left(\frac{\pi}{2}-\left(\frac{A+B}{2}\right)\right)\right. \left. \cos \left(\frac{\pi}{2}-\left(\frac{A+C}{2}\right)\right)\right] \\ & =4 \sin \left(\frac{A+B}{2}\right) \sin \left(\frac{A+C}{2}\right) \sin \left(\frac{A+D}{2}\right) \end{aligned} $

$ \begin{aligned} &\text { } \cos x+\sin x=\frac{1}{2}, 0 < x < \pi\\ &\begin{aligned} \Rightarrow & \quad(\cos x+\sin x)^2=\frac{1}{4} \\ \Rightarrow & \quad 1+\sin 2 x=\frac{1}{4} \\ \Rightarrow & \quad \sin 2 x=\frac{-3}{4} \\ \Rightarrow & \quad \frac{2 \tan x}{1+\tan ^2 x}=\frac{-3}{4} \\ \Rightarrow & \quad 8 \tan x=-3-3 \tan ^2 x \\ \Rightarrow \quad & \quad 3 \tan ^2 x+8 \tan x+3=0 \\ \Rightarrow \quad & \tan x=\frac{-8 \pm \sqrt{64-36}}{6} \\ & =\frac{-8 \pm \sqrt{28}}{6} \\ & =\frac{-4 \pm \sqrt{7}}{3} \\ & =\frac{-4-\sqrt{7}}{3}, \frac{-4+\sqrt{7}}{3} \end{aligned} \end{aligned} $

Given, $\sin \theta+2 \cos \theta=1$

$ \begin{aligned} \Rightarrow & 2 \cos \theta=1-\sin \theta \\ \Rightarrow & 4 \cos ^2 \theta=1+\sin ^2 \theta-2 \sin \theta \\ \Rightarrow & \sin ^2 \theta+1-4\left(1-\sin ^2 \theta\right)-2 \sin \theta=0 \\ \Rightarrow & 5 \sin ^2 \theta-2 \sin \theta-3=0 \\ \Rightarrow & 5 \sin ^2 \theta-5 \sin \theta+3 \sin \theta-3=0 \\ \Rightarrow & 5 \sin \theta(\sin \theta-1)+3(\sin \theta-1)=0 \\ & (\sin \theta-1)(5 \sin \theta+3)=0 \\ & \sin \theta=1, \quad \sin \theta=\frac{-3}{5} \end{aligned} $

$\because \theta$ lies in 4th quadrant

$ \begin{aligned} & \therefore \sin \theta=\frac{-3}{5} \text { and } \cos \theta=\frac{4}{5} \\ & \therefore 7 \cos \theta+6 \sin \theta=\frac{28}{5}-\frac{18}{5}=\frac{10}{5}=2 \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { We have, } 3 \cos ^2 A+2 \cos ^2 B=4 \\ & \Rightarrow \quad \frac{3(1+\cos 2 A)}{2}+\frac{2(1+\cos 2 \beta)}{2}=4 \\ & \Rightarrow \quad 3 \cos 2 A+2 \cos 2 \beta=3 \\ & \Rightarrow \quad 2 \cos 2 \beta=3(1-\cos 2 A)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \\ & \text { and } \frac{3 \sin A}{\sin B}=\frac{2 \cos B}{\cos A} \\ & \Rightarrow \quad 3 \sin A \cos A=2 \sin B \cos B \\ & \Rightarrow \quad \frac{3}{2} \sin 2 A=\sin 2 B \\ & \Rightarrow \quad 3 \sin 2 A=2 \sin 2 B \end{aligned} \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad 9 \sin ^2 2 A=4 \sin ^2 2 B \\ & \Rightarrow \quad 9-9 \cos ^2 2 A=4-4 \cos ^2 2 B \\ & 9 \cos ^2 2 A-4 \cos ^2 2 B=5 \end{aligned} $

From Eq. (i)

$ \begin{aligned} & \Rightarrow 9 \cos ^2 2 A-9(1-\cos 2 A)^2=5 \\ & \Rightarrow 9 \cos ^2 2 A-9 \cos ^2 2 A-9+18 \cos 2 A=5 \\ & \Rightarrow 18 \cos 2 A=14 \\ & \Rightarrow \cos 2 A=\frac{14}{18}=\frac{7}{9} \end{aligned} $

From Eq. (i), we get

$ \begin{aligned} \therefore \cos 2 B & =\frac{3}{2}\left(1-\frac{14}{18}\right), \cos A=\sqrt{\frac{1+\cos 2 A}{2}} \\ & =\frac{3}{2} \times \frac{4}{18}=\frac{1}{3} \quad=\sqrt{\frac{16}{18}}=\frac{2 \sqrt{2}}{3} \end{aligned} $

Now,

$ \begin{aligned} \sin (A+2 B) & =\sin A \cos 2 B+\cos A \sin 2 B \\ & =\frac{1}{3} \times \frac{1}{3}+\frac{2 \sqrt{2}}{3} \times \frac{2 \sqrt{2}}{3} \\ & =\frac{1}{9}+\frac{8}{9}=1 \\ \therefore A+2 B & =\frac{\pi}{2} \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} & \left(4 \cos ^2 \frac{\pi}{20}-1\right)\left(4 \cos ^2 \frac{3 \pi}{20}-1\right) \\ & \left(4 \cos ^2 \frac{5 \pi}{20}+1\right)\left(4 \cos ^2 \frac{7 \pi}{20}-1\right) \left(4 \cos ^2 \frac{9 \pi}{20}-1\right) \end{aligned} \end{aligned} $

$ \begin{aligned} & =\left(4 \cos ^2 \frac{\pi}{20}-1\right)\left(4 \cos ^2 \frac{3 \pi}{20}-1\right)\left(4 \times \frac{1}{2}+1\right) \\ & \left(4 \cos ^2 \frac{7 \pi}{20}-1\right)\left(4 \cos ^2 \frac{9 \pi}{20}-1\right) \\ & =3\left(4 \cos ^2 \frac{\pi}{20}-1\right)\left(4 \cos ^2 \frac{9 \pi}{20}-1\right) \\ & \left(4 \cos ^2 \frac{3 \pi}{20}-1\right)\left(4 \cos ^2 \frac{7 \pi}{20}-1\right) \\ & =3\left(4 \cos ^2 \frac{\pi}{20}-1\right)\left(4 \sin ^2 \frac{\pi}{20}-1\right) \\ & \left(4 \cos ^2 \frac{3 \pi}{20}-1\right)\left(4 \sin ^2 \frac{3 \pi}{20}-1\right) \\ & =3\left(16 \cos ^2 \frac{\pi}{20} \cdot \sin ^2 \frac{\pi}{20}-4+1\right) \left(16 \sin ^2 \frac{3 \pi}{20} \cos ^2 \frac{3 \pi}{20}-4-1\right) \end{aligned} $

$ \begin{aligned} & =3\left(4 \sin ^2 \frac{\pi}{10}-3\right)\left(4 \sin ^2 \frac{3 \pi}{10}-3\right) \\ & =\frac{3}{\sin \frac{\pi}{10} \cdot \sin \frac{3 \pi}{10}}\left(4 \sin ^3 \frac{\pi}{10}-3 \sin \frac{\pi}{10}\right) \left(4 \sin ^3 \frac{3 \pi}{10}-3 \sin \frac{3 \pi}{10}\right) \end{aligned} $

$ \begin{aligned} & =\frac{3}{\sin \frac{\pi}{10} \sin \frac{3 \pi}{10}} \cdot \sin \frac{3 \pi}{10} \cdot \sin \left(\frac{9 \pi}{10}\right) \\ & =3 \quad\left(\because \sin \left(\frac{9 \pi}{10}\right)=\sin \left(\pi-\frac{\pi}{10}\right)=\sin \frac{\pi}{10}\right) \end{aligned} $

$ \begin{aligned} &\text {Given, }\\ &\begin{aligned} & 2 \tan (A+B)=3 \tan (A-B) \\ & \Rightarrow \frac{\tan (A+B)}{\tan (A-B)}=\frac{3}{2} \\ & \Rightarrow \frac{\tan (A+B)+\tan (A-B)}{\tan (A+B)-\tan (A-B)}=\frac{3+2}{3-2} \\ & \Rightarrow \frac{[\sin (A+B) \cos (A-B)}{[\sin (A+B) \cos (A-B)} -\sin (A-B) \cos (A+B)] \\ & \Rightarrow \frac{\sin (A+B+A-B)}{\sin (A+B-A+B)}=5 \\ & \Rightarrow \sin 2 A=5 \sin 2 B \\ & \Rightarrow \sin A \cos A=5 \sin B \cos B \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} & \cos ^3 80^{\circ}+\cos ^3 40^{\circ}-\cos ^3 20^{\circ}=k \\ & \left(\cos 80^{\circ}+\cos 40^{\circ}\right)\left(\left(\cos 80^{\circ}+\cos 40^{\circ}\right)^2\right. \left.-3 \cos 80^{\circ} \cos 40^{\circ}\right)-\cos ^3 20^{\circ}=k \\ & \Rightarrow \quad 2 \cos 60^{\circ} \cos 20^{\circ}\left(\left(2 \cos 60^{\circ} \cos 20^{\circ}\right)^2\right. \left.-\frac{3}{2}\left(\cos 120^{\circ}+\cos 40^{\circ}\right)\right]-\cos ^3 20^{\circ}=k \\ & \Rightarrow \quad \cos 20^{\circ}\left(\cos ^2 20^{\circ}-\frac{3}{2}\left(-\frac{1}{2}+\cos 40^{\circ}\right)\right. \left.-\cos ^3 20^{\circ}=k\right) \\ & \Rightarrow \quad \cos ^3 20^{\circ}-\frac{3}{2} \cos 20^{\circ}\left(-\frac{1}{2}+\cos 40^{\circ}\right) -\cos ^3 20^{\circ}=k \end{aligned} \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad \frac{3}{4} \cos 20^{\circ}-\frac{3}{2} \cos 20^{\circ} \cos 40^{\circ}=k \\ & \Rightarrow \quad \frac{3}{4} \cos 20^{\circ}-\frac{3}{4}\left(\cos 60^{\circ}+\cos 20^{\circ}\right)=k \\ & \Rightarrow \quad \frac{3}{4} \cos 20-\frac{3}{4} \times \frac{1}{2}-\frac{3}{4} \cos 20^{\circ}=k \\ & \Rightarrow \quad k=-\frac{3}{8} \Rightarrow \frac{4 k}{3}=\frac{-1}{2}=\cos \frac{2 \pi}{3} \end{aligned} $

$ \begin{aligned} & \text { } \begin{array}{l} \cos 13^{\circ} \cdot \sin 17^{\circ} \cdot \sin 21^{\circ} \cdot \cos 47^{\circ} \\ =\left(\cos 13^{\circ} \cos 47^{\circ}\right) \cdot\left(\sin 17^{\circ} \sin 21^{\circ}\right) \\ =\frac{1}{2}\left[\cos \left(34^{\circ}\right)+\cos \left(60^{\circ}\right)\right] \cdot \frac{1}{2}\left[\cos \left(4^{\circ}\right)-\cos \left(38^{\circ}\right)\right] \\ \quad\left[\because \sin A \sin B=\frac{1}{2}[\cos (A-B)-\cos (A+B)\right. \text { and } \cos A \cos B=\frac{1}{2}[\cos (A-B)+\cos (A+B)] \\ =\frac{1}{4}\left[\cos 34^{\circ}+\frac{1}{2}\right]\left[\cos 4^{\circ}-\cos 38^{\circ}\right] \\ =\frac{1}{4}\left[\cos 34^{\circ} \cos 4^{\circ}-\cos 34^{\circ} \cos 38^{\circ}+\frac{1}{2} \cos 4^{\circ}-\frac{1}{2} \cos 38^{\circ}\right] \end{array} \end{aligned} $

$ \begin{aligned} & =\frac{1}{4}\left[\frac{1}{2}\left\{\cos \left(30^{\circ}\right)+\cos \left(38^{\circ}\right)\right\}-\frac{1}{2}\left\{\cos \left(4^{\circ}\right)+\cos \left(72^{\circ}\right)\right\}\right. \left.+\frac{1}{2} \cos 4^{\circ}-\frac{1}{2} \cos 38^{\circ}\right] \\ & =\frac{1}{4}\left[\frac{1}{2}\left(\frac{\sqrt{3}}{2}+\cos 38^{\circ}\right)-\frac{1}{2}\left(\cos 4^{\circ}+\cos 72^{\circ}\right)+\frac{1}{2} \cos 4^{\circ}-\frac{1}{2} \cos 38^{\circ}\right] \\ & =\frac{1}{8}\left[\frac{\sqrt{3}}{2}-\cos 72^{\circ}\right]=\frac{\sqrt{3}}{16}-\frac{\cos 72^{\circ}}{8} \\ & =\frac{\sqrt{3}}{16}-\frac{\sqrt{5}-1}{4} \times \frac{1}{8} \left(\because \cos 72^{\circ}=\frac{\sqrt{5}-1}{4}\right)\\ & =\frac{\sqrt{3}}{16}-\frac{\sqrt{5}-1}{32}=\frac{2 \sqrt{3}-\sqrt{5}+1}{32}=\frac{1}{32}(2 \sqrt{3}-\sqrt{5}+1) \end{aligned} $

$ \begin{aligned} & \sin \frac{\pi}{5}+\sin \frac{2 \pi}{5}+\sin \frac{3 \pi}{5}+\sin \frac{4 \pi}{5} \\ & =\sin \frac{\pi}{5}+\sin \frac{2 \pi}{5}+\sin \left(\pi-\frac{2 \pi}{5}\right)+\sin \left(\pi-\frac{\pi}{5}\right) \\ & =\sin \frac{\pi}{5}+\sin \frac{2 \pi}{5}+\sin \frac{2 \pi}{5}+\sin \frac{\pi}{5}=2 \sin \frac{\pi}{5}+2 \sin \frac{2 \pi}{5} \\ & =2\left[\sin \frac{\pi}{5}+\sin \frac{2 \pi}{5}\right]=4 \sin \frac{3 \pi}{10} \cos \frac{\pi}{10} \\ & =4 \frac{(\sqrt{5}+1)}{4} \frac{(\sqrt{10+2 \sqrt{5}})}{4}=\frac{1}{4}(\sqrt{5}+1)(\sqrt{10+2 \sqrt{5}}) \end{aligned} $

$ \begin{aligned} \operatorname{cosec} 48^{\circ}+\operatorname{cosec} 96^{\circ} & +\operatorname{cosec} 192^{\circ} +\operatorname{cosec} 384^{\circ} \end{aligned} $

$ \begin{aligned} \because \quad \operatorname{cosec} \theta & =\cot \theta / 2-\cot \theta \\ \therefore \operatorname{cosec} 48^{\circ} & =\cot 24^{\circ}-\cot 48^{\circ} \\ \operatorname{cosec} 96^{\circ} & =\cot 48^{\circ}-\cot 96^{\circ} \\ \operatorname{cosec} 192^{\circ} & =\cot 96^{\circ}-\cot 192^{\circ} \\ \operatorname{cosec} 384^{\circ} & =\operatorname{cosec} 24^{\circ} \\ =\cot 24^{\circ} & -\cot 48^{\circ}+\cot 48^{\circ}-\cot 96^{\circ}+\cot 96^{\circ}-\cot 192^{\circ}+\cos 24^{\circ} \end{aligned} $

$ \begin{aligned} &=\cot 24^{\circ}-\cot 192^{\circ}+\cos 24^{\circ}\\ &\text { and } \operatorname{cosec} 24^{\circ}=\cot 12^{\circ}-\cot 24^{\circ}\\ &\begin{aligned} & =\cot 24^{\circ}-\cot 192^{\circ}+\cot 12^{\circ}-\cot 24^{\circ} \\ & =\cot 12^{\circ}-\cot 192^{\circ} \\ & =\cot 12^{\circ}-\cot \left(180+12^{\circ}\right) \\ & =\cot 12^{\circ}-\cot 12^{\circ}=0 \end{aligned} \end{aligned} $

$\because \cos \theta=-\frac{3}{5}<0$ and $\theta$ does not lie in second quadrant.

Thus, $\theta$ lies is third quadrant.

So, $\theta / 2$ must be lie in second quadrant.

There $\tan \theta / 2$ must be negative.

then, $\cos \theta=\frac{3}{5}=\frac{B}{H}$

$\therefore \quad \tan \theta=\frac{4}{3}$

So, $\tan \theta=\frac{2 \tan \theta / 2}{1-\tan ^2 \theta / 2}$

let $\tan \theta / 2=x$

$ \begin{aligned} & \Rightarrow \quad \frac{4}{3}=\frac{2 x}{1-x^2} \Rightarrow 4-4 x^2=6 x \\ & \Rightarrow \quad 2 x^2+3 x-2=0 \\ & \Rightarrow \quad 2 x^2+4 x-x-2=0 \\ & \Rightarrow \quad(2 x-1)(x+2)=0 \end{aligned} $

Therefore,

$ \begin{aligned} & x=1 / 2 \text { or } x=-2 \\ \Rightarrow \quad & \tan \theta / 2=1 / 2 \text { or } \tan \theta / 2=-2 \end{aligned} $

$\because \tan \theta / 2$ is negative

Hence, $\tan \theta / 2=-2$

$ \begin{aligned} &\\ &\begin{aligned} & \text { } \begin{aligned} Let\,\,f(x) & =\cos ^2\left(\frac{x}{4}\right)+\sin \left(\frac{x}{4}\right) \\ & =1-\sin ^2\left(\frac{x}{4}\right)+\sin \left(\frac{x}{4}\right) \\ & =-\sin ^2\left(\frac{x}{4}\right)+\sin \left(\frac{x}{4}\right)+1 \\ & =\frac{5}{4}-\left(\sin \left(\frac{x}{4}\right)-\frac{1}{2}\right)^2 \end{aligned} \\ & \left.\therefore f(x)\right|_{\max }=\alpha=\frac{5}{4} \text { when } \sin \frac{x}{4}=\frac{1}{2} \\ & \text { and }\left.f(x)\right|_{\min }=\beta=-1 \text { when } \sin \frac{x}{4}=-1 \\ & \therefore \alpha-\beta=\frac{5}{4}-(-1)=\frac{9}{4} \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { } 3 \cos ^2 A+2 \cos ^2 B=4 \\ & \Rightarrow 3\left(1-\sin ^2 A\right)+2\left(1-\sin ^2 B\right)=4 \end{aligned} $

Now, $3 \sin A \cdot \cos A=2 \sin B \cdot \cos B$ (given)

$ \Rightarrow \quad 3 \sin 2 A=2 \sin 2 B \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

$ \begin{aligned} & \cos (A+2 B)=\cos A \cos 2 B-\sin A \sin 2 B \\ & =\cos A \cdot 3 \sin ^2 A-\sin A \cdot \frac{3}{2} \cdot 2 \sin A \cos A \,\,\,\,(from Eq. (i))\text { } \end{aligned} $

$ =3 \cos A \sin ^2 A-3 \cos A \sin ^2 A=0 $

$ \therefore A+2 B=90^{\circ} $

$ \begin{aligned} &\text { Given, } \sin x-\sin y=27 / 65 \text { and }\\ &\begin{aligned} & \cos x-\cos y=\frac{-21}{65} \\ & 2 \sin \left(\frac{x-y}{2}\right) \cos \left(\frac{x+y}{2}\right)=\frac{27}{65} \text { and } \\ & 2 \sin \left(\frac{x+y}{2}\right) \sin \left(\frac{x-y}{2}\right)=\frac{21}{65} \\ & \therefore \tan \left(\frac{x+y}{2}\right)=\frac{21}{27}=\frac{7}{9} \\ & \sin (x+y)=\frac{2 \tan \left(\frac{x+y}{2}\right)}{1+\tan ^2 \frac{x+y}{2}}=\frac{2 \times \frac{7}{9}}{1+\frac{49}{81}} \\ & \sin (x+y)=\frac{63}{65} \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { } \sin \beta=\frac{5}{6} \sin \alpha\\ &\begin{aligned} & \Rightarrow \cos \beta=\frac{5 \sqrt{5}}{9} \cos \alpha \\ & \Rightarrow \sin ^2 \beta+\cos ^2 \beta=1 \\ & \Rightarrow\left(\frac{5}{6} \sin \alpha\right)^2+\left(\frac{5 \sqrt{5}}{9} \cos \alpha\right)^2=1 \\ & \Rightarrow \frac{25}{36} \sin ^2 \alpha+\frac{125}{81} \cos ^2 \alpha=1 \\ & \Rightarrow \frac{25}{36} \sin ^2 \alpha+\frac{125}{81}-\frac{125}{81} \sin ^2 \alpha=1 \\ & \Rightarrow\left(\frac{225-500}{324}\right) \sin ^2 \alpha=\frac{81-125}{81} \\ & \Rightarrow \sin ^2 \alpha=\frac{-44}{81} \times \frac{324}{-275} \\ & \Rightarrow \sin ^2 \alpha=\frac{4}{9} \times \frac{11}{9} \times \frac{4}{25} \times \frac{81}{11} \\ & \Rightarrow \sin ^2 \alpha=\frac{16}{25} \Rightarrow \sin \alpha=\frac{4}{5} \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { }\left(\frac{\sin 3 \theta}{\sin \theta}\right)^2-\left(\frac{\cos 3 \theta}{\cos \theta}\right)^2=a \cos b \theta \\ & \Rightarrow\left(\frac{3 \sin \theta-4 \sin ^3 \theta}{\sin \theta}\right)-\left(\frac{4 \cos ^3 \theta-3 \cos \theta}{\cos \theta}\right) =a \cos b \theta \\ & \Rightarrow\left(3-4 \sin ^2 \theta\right)^2-\left(4 \cos ^2 \theta-3\right)^2 =a \cos b \theta \\ & \Rightarrow 9+16 \sin ^4 \theta-24 \sin ^2 \theta-16 \cos ^4 \theta -9+24 \cos ^2 \theta=a \cos b \theta \\ & \Rightarrow 24\left(\cos ^2 \theta-\sin ^2 \theta\right)-16\left(\cos ^4 \theta-\sin ^4 \theta\right) =a \cos ^b \theta \\ & \Rightarrow 8\left(\cos ^2 \theta-\sin ^2 \theta\right)\left(3-2\left(\cos ^2 \theta+\sin ^2 \theta\right)\right) =a \cos b \theta \quad, \\ & \Rightarrow 8 \cos 2 \theta(3-2)=a \cos b \theta \\ & \Rightarrow 8 \cos 2 \theta=a \cos b \theta \Rightarrow a=8, b=2 \\ & \therefore a: b=8: 2=4: 1 \end{aligned} $

Let $x_1$ be the initial horizontal distance

$ \tan 15^{\circ}=\frac{5}{x_1} \Rightarrow x_1=\frac{5}{\tan 15^{\circ}} $

Let $x_2$ be the final horizontal distance

$ \begin{array}{rlrl} \tan 30^{\circ} & =\frac{5}{x_2} \\ \Rightarrow & \quad x_2 & =\frac{5}{\tan 30^{\circ}} \end{array} $

The distance travelled is $d=x_1-x_2$

$ \begin{aligned} & d=\frac{5}{\tan 15^{\circ}}-\frac{5}{\tan 30^{\circ}} \\ & d \approx 5(3.732-1.732) \\ & d \approx 5 \times 2 \\ & d=10 \mathrm{~km} \end{aligned} $

$ \begin{aligned} &\begin{aligned} \text { speed } & =\frac{d}{t}=\frac{10}{50}=0.2 \mathrm{~km} / \mathrm{s} \\ & =3600 \times 0.2 \mathrm{~km} / \mathrm{h}=720 \mathrm{~km} / \mathrm{h} \end{aligned}\\ &\Rightarrow \text { The speed of the plane is } 720 \mathrm{~km} / \mathrm{h} \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} A+B= & \frac{\pi}{4} \\ \Rightarrow \frac{\cos B-\sin B}{\cos B+\sin B} & =\frac{\sin \left(\frac{\pi}{4}+A\right)-\sin \left(\frac{\pi}{4}-A\right)}{\sin \left(\frac{\pi}{4}+A\right)+\sin \left(\frac{\pi}{4}-A\right)} \\ & =\frac{2 \cos \frac{\pi}{4} \sin A}{2 \sin \frac{\pi}{4} \cos A}=\tan A \end{aligned} \end{aligned} $

$7 \cos \theta-\sin \theta=5, \tan \theta>0$

$ 7-\tan \theta=5 \sec \theta $

Squaring both sides, we get

$ \begin{aligned} & \Rightarrow 49+\tan ^2 \theta-14 \tan \theta=25\left(1+\tan ^2 \theta\right) \\ & \Rightarrow 24 \tan ^2 \theta+14 \tan \theta-24=0 \\ & \Rightarrow 12 \tan ^2 \theta+7 \tan \theta-12=0 \\ & \Rightarrow 12 \tan ^2 \theta+16 \tan \theta-9 \cdot \tan \theta-12=0 \\ & \Rightarrow(3 \tan \theta+4)(4 \tan \theta-3)=0 \\ & \quad \tan \theta=\frac{3}{4} \end{aligned} $

$ \begin{aligned} &\text { } \sin ^3 10^{\circ}+\sin ^3 50^{\circ}-\sin ^3 70^{\circ}\\ &\begin{aligned} & \Rightarrow \quad \sin 3 x=3 \sin x-4 \sin ^3 x \\ & \Rightarrow \quad \sin ^3 x=\frac{1}{4}(3 \sin x-\sin 3 x) \\ & \Rightarrow \quad \frac{1}{4}\left[3 \sin 10^{\circ}-\sin 30^{\circ}+3 \sin 50^{\circ}\right. \left.-\sin 150^{\circ}-3 \sin 70^{\circ}+\sin 210^{\circ}\right] \\ & \Rightarrow \quad \frac{1}{4}\left[3 \sin 10^{\circ}+3 \sin 50^{\circ}-3 \sin 70^{\circ}-\frac{1}{2}-\frac{1}{2}-\frac{1}{2}\right] \\ & \Rightarrow \frac{3}{4}\left[\sin 10^{\circ}+\sin 50^{\circ}-\sin 70^{\circ}-\frac{1}{2}\right] \\ & \Rightarrow \frac{3}{4}\left[2 \sin 30^{\circ} \cos 20^{\circ}-\sin 70^{\circ}-\frac{1}{2}\right] \\ & \Rightarrow \frac{3}{4}\left[\cos 20^{\circ}-\sin 70^{\circ}-\frac{1}{2}\right] \\ & \Rightarrow \frac{3}{4}\left[\sin 70^{\circ}-\sin 70^{\circ}-\frac{1}{2}\right]=-\frac{3}{8} \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { We have, } \sum_{r=1}^{89} \frac{1}{\sin r^{\circ} \cdot \sin (r+1)^{\circ}}\\ &\begin{aligned} & \Rightarrow \frac{1}{\sin 1^{\circ}} \sum_{r=1}^{89} \frac{\sin \left(\left(r^{\circ}+1^{\circ}\right)-r^{\circ}\right)}{\sin r^{\circ} \cdot \sin \left(r^{\circ}+1^{\circ}\right)} \\ & \Rightarrow \frac{1}{\sin 1^{\circ}} \sum_{r=1}^{89} \frac{\sin (r+1)^{\circ} \cos r^{\circ}}{\sin r^{\circ} \cdot \sin (r+1)^{\circ}} \\ & \Rightarrow \frac{1}{\sin 1^{\circ}} \sum_{r=1^{\circ}}^{89} \cot r^{\circ}-\cot (r+1)^{\circ} \\ & \Rightarrow \frac{1}{\sin 1^{\circ}}\left[\cot 1^{\circ}-\cot 90^{\circ}\right]=\frac{\cot 1^{\circ}}{\sin 1^{\circ}} \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { Given, }\\ &\begin{aligned} & \cos ^3 \frac{\pi}{8} \cdot \cos \frac{3 \pi}{8}+\sin ^3 \frac{\pi}{8} \cdot \sin \frac{3 \pi}{8} \\ & \text { Using } \frac{3 \pi}{8}=\frac{\pi}{2}-\frac{\pi}{8} \\ & \Rightarrow \cos \frac{3 \pi}{8}=\sin \frac{\pi}{8} \text { and } \sin \frac{3 \pi}{8}=\cos \frac{\pi}{8} \\ & \Rightarrow \cos ^3 \frac{\pi}{8} \cdot \sin \frac{\pi}{8}+\sin ^3 \frac{\pi}{8} \cdot \cos \frac{\pi}{8} \\ & \quad=\sin \frac{\pi}{8} \cdot \cos \frac{\pi}{8}\left(\cos ^2 \frac{\pi}{8}+\sin ^2 \frac{\pi}{8}\right) \\ & \quad=\frac{1}{2} 2 \sin \frac{\pi}{8} \cos \frac{\pi}{8} \\ & \quad=\frac{1}{2} \sin \frac{\pi}{4}=\frac{1}{2 \sqrt{2}} \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { Given, } A+B+C=\frac{\pi}{4}\\ &\begin{aligned} & \therefore 2 A+2 B+2 C=\frac{\pi}{2} \\ & \text { Now, } \sin 4 A+\sin 4 B+\sin 4 C \\ & 2 \sin (2 A+2 B) \cdot \cos (2 A-2 B)+2 \sin 2 C \\ & =2 \sin \left(\frac{\pi}{2}-2 C\right) \cdot \cos 2 C +2 \sin 2 C \cdot \cos 2 C \\ & =2 \cos 2 C \cdot \cos (2 A-2 B)+2 \sin 2 C \cdot \cos 2 C \\ & =2 \cos 2 C\{\cos (2 A-2 B)+\sin 2 C\} \\ & =2 \cos 2 C[\cos (2 A-2 B)+\cos (2 A+2 B)] \\ & =2 \cos 2 C[2 \cos 2 A \cdot \cos 2 B] \\ & =4 \cos 2 A \cdot \cos 2 B \cdot \cos 2 C \end{aligned} \end{aligned} $

Given, $630^{\circ}<\theta<810^{\circ}$ and

$ \tan \theta=\frac{-7}{24} $

$\therefore \theta$ lies IV quadrant

$ \begin{aligned} \therefore \quad \cos \theta & =\frac{24}{25} \quad 315^{\circ}<\frac{\theta}{2}<405^{\circ} \\ 2 \cos ^2 \frac{\theta}{2}-1 & =\frac{24}{25} \quad 157 \frac{1}{2}<\frac{\theta}{4}<2.02 \frac{1}{2} \\ 2 \cos ^2 \frac{\theta}{2} & =\frac{49}{25} \quad \frac{\theta}{4} \text { lies III quadrant } \\ \Rightarrow \quad \cos \frac{\theta}{2} & =\frac{7}{5 \sqrt{2}} \Rightarrow 2 \cos ^2 \frac{\theta}{4}-1=\frac{7}{5 \sqrt{2}} \\ \Rightarrow 2 \cos ^2 \frac{\theta}{4} & =1+\frac{7}{5 \sqrt{2}}=\frac{7+5 \sqrt{2}}{5 \sqrt{2}} \\ \Rightarrow \quad \cos \frac{\theta}{4} & =-\sqrt{\frac{7+5 \sqrt{2}}{10 \sqrt{2}}} \end{aligned} $

We have,

$ 2 \cos \theta+\sin \theta=1 \quad \theta \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] $

Squaring both sides, we get

$ \begin{aligned} & \Rightarrow \quad(2 \cos \theta)^2=(1-\sin \theta)^2 \\ & \Rightarrow \quad 4 \cos ^2 \theta=1+\sin ^2 \theta-2 \sin \theta \\ & \Rightarrow \quad 5 \sin ^2 \theta-2 \sin \theta-3=0 \\ & \Rightarrow \quad(\sin \theta-1)(5 \sin \theta+3)=0 \\ & \Rightarrow \quad \sin \theta=1 \text { and }-\frac{3}{5} \\ & \Rightarrow \quad \cos \theta=0 \text { and }+\frac{4}{5} \\ & \because \quad 7 \cos \theta+6 \sin \theta=k, \text { when } \sin \theta=1 \\ & \Rightarrow \quad 7(0)+6(1)=k \Rightarrow k=6 \end{aligned} $

$ \begin{aligned} &\text { When, } \sin \theta=-\frac{3}{5} \text { and } \cos \theta=\frac{4}{5}\\ &k=7\left(\frac{4}{5}\right)-6\left(\frac{3}{5}\right)=\frac{28-18}{5}=\frac{10}{5}=2 \end{aligned} $

$ \begin{aligned} &\sum_{k=0}^{12} \frac{1}{\sin \left((k+1) \frac{\pi}{6}+\frac{\pi}{4}\right) \cdot \sin \left(\frac{k \pi}{6}+\frac{\pi}{4}\right)}\\ &\text { Multiplying } N^r \text { and } D^r \text { by } \sin \pi / 6 \text {, }\\ &\begin{aligned} & \frac{1}{\sin \frac{\pi}{6}} \sum_{k=0}^{12} \frac{\sin \left\{\left((k+1) \frac{\pi}{6}+\frac{4}{\pi}\right)-\left(\frac{k \pi}{6}+\frac{\pi}{4}\right)\right\}}{2 \sum_{k=0}^{12}\left((k+1) \frac{\pi}{6}+\frac{\pi}{4}\right) \cdot \sin \left(\frac{k \pi}{6}+\frac{\pi}{4}\right)} \\ & =2\left[\cot \left(\frac{k \pi}{6}+\frac{\pi}{4}\right)-\cot \left((k+1) \frac{\pi}{6}+\frac{\pi}{4}\right)\right\} \\ & \quad+\left\{\cot \left(\frac{\pi}{6}+\frac{\pi}{4}\right)-\cot \left(\frac{\pi}{6}+\frac{\pi}{4}\right)\right\} \\ & \quad+\ldots+\left\{\cot \left(\frac{12 \pi}{6}+\frac{\pi}{4}\right)-\cot \left(\frac{13 \pi}{6}+\frac{\pi}{4}\right)\right\} \\ & =2\left[\cot \frac{\pi}{4}+\cot \left(\frac{\pi}{6}+\frac{\pi}{4}\right)\right] \\ & =2\left[1-\frac{\sqrt{3}-1}{\sqrt{3}+1}\right]=2\left(\frac{2}{\sqrt{3}+1}\right)=\frac{4(\sqrt{3}-1)}{2} \\ & =2(\sqrt{3}-1) \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { Given } \cos \alpha=\sec h \beta\\ &\begin{aligned} & \text { or } \quad \cos h \beta=\sec \alpha \\ & \Rightarrow \quad \beta=\cos h^{-1}(\sec \alpha) \\ & \Rightarrow \quad \beta=\log \left(\sec \alpha+\sqrt{\sec ^2 \alpha-1}\right) \\ & {\left[\because \cos h^{-1} x=\log \left(x+\sqrt{x^2-1}\right)\right]} \\ & \beta=\log (\sec \alpha+\tan \alpha) \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { } \tan ^2 \frac{\pi}{16}+\tan ^2 \frac{2 \pi}{16}+\tan ^2 \frac{3 \pi}{16} \\ & \begin{array}{r} +\tan ^2 \frac{4 \pi}{16}+\tan ^2 \frac{5 \pi}{16}+\tan ^2 \frac{6 \pi}{16} \\ +\tan ^2 \frac{7 \pi}{16}=\left(\sec ^2 \frac{\pi}{16}-1\right)+\left(\sec ^2 \frac{2 \pi}{16}-1\right) \\ +\left(\sec ^2 \frac{3 \pi}{16}-1\right)+1+\tan ^2\left(\frac{\pi}{2}-\frac{3 \pi}{16}\right) \\ +\tan ^2\left(\frac{\pi}{2}-\frac{2 \pi}{16}\right)+\tan ^2\left(\frac{\pi}{2}-\frac{\pi}{16}\right) \end{array} \\ & \begin{array}{r} \left(\sec ^2 \frac{\pi}{16}-1\right)+\left(\sec ^2 \frac{2 \pi}{16}-1\right) +\left(\sec ^2 \frac{3 \pi}{16}-1\right)+1 \end{array} \end{aligned} $

$ \begin{aligned} & +\cot ^2\left(\frac{3 \pi}{16}\right)+\cot ^2\left(\frac{2 \pi}{16}\right)+\cot ^2\left(\frac{\pi}{16}\right) \\ & =\left(\sec ^2 \frac{\pi}{16}-1\right)+\left(\sec ^2 \frac{2 \pi}{16}-1\right) \\ & +\left(\sec ^2 \frac{3 \pi}{16}-1\right)+1+\left(\operatorname{cosec}^2 \frac{3 \pi}{16}-1\right) \\ & +\left(\operatorname{cosec}^2 \frac{2 \pi}{16}-1\right)+\left(\operatorname{cosec}^2 \frac{\pi}{16}-1\right) \\ & =\left(\sec ^2 \frac{\pi}{16}+\operatorname{cosec}^2 \frac{\pi}{16}\right) +\left(\sec ^2 \frac{2 \pi}{16}+\operatorname{cosec}^2 \frac{2 \pi}{16}\right) +\left(\sec ^2 \frac{3 \pi}{16}+\operatorname{cosec}^2 \frac{3 \pi}{16}\right)-6+1 \end{aligned} $

$ \begin{aligned} & =\frac{1}{\sin ^2 \frac{\pi}{16} \cos ^2 \frac{\pi}{16}}+\frac{1}{\sin ^2 \frac{2 \pi}{16} \cos ^2 \frac{2 \pi}{16}} \\ & +\frac{1}{\sin ^2 \frac{3 \pi}{16} \cos ^2 \frac{3 \pi}{16}}-5 \\ & =\frac{4}{\sin ^2 \frac{\pi}{8}}+\frac{4}{\sin ^2 \frac{\pi}{4}}+\frac{4}{\sin ^2 \frac{3 \pi}{8}}-5 \\ & =4\left(\frac{1}{\sin ^2 \frac{\pi}{8}}+\frac{1}{\cos ^2 \frac{\pi}{8}}\right)+4 \times 2-5 \\ & =4\left(\frac{1}{\sin ^2 \frac{\pi}{8} \cos ^2 \frac{\pi}{8}}\right)+3=\frac{4 \times 4}{\sin ^2 \frac{\pi}{4}}+3 \\ & =2 \times 16+3=35 \end{aligned} $

$ \begin{aligned} & \text { } \sin ^2 18^{\circ}+\sin ^2 24^{\circ}+\sin ^2 36^{\circ} +\sin ^2 42^{\circ}+\sin ^2 78^{\circ}+\sin ^2 90^{\circ} \\ & +\sin ^2 96^{\circ}+\sin ^2 102^{\circ}+\sin ^2 138^{\circ} +\sin ^2 162^{\circ} \\ & \because \sin ^2 \theta=\sin ^2\left(180^{\circ}-\theta\right) \end{aligned} $

$ \begin{aligned} & \begin{array}{r} \therefore \sin ^2 18^{\circ}+\sin ^2 24^{\circ}+\sin ^2 36^{\circ}+\sin ^2 42^{\circ} +\sin ^2 78^{\circ}+1+\sin ^2 84^{\circ} +\sin ^2 78^{\circ}+\sin ^2 42^{\circ}+\sin ^2 18^{\circ} \end{array} \\ & \begin{array}{r} =2 \sin ^2 18^{\circ}+2 \sin ^2 42^{\circ}+2 \sin ^2 78^{\circ} +\sin ^2 24^{\circ}+\sin ^2 36^{\circ}+\sin ^2 84^{\circ}+1 \end{array} \\ & \begin{array}{r} =2 \sin ^2 18^{\circ}+2 \sin ^2 42^{\circ}+2 \cos ^2 12^{\circ} +\sin ^2 24^{\circ}+\sin ^2 36^{\circ}+\cos ^2 6^{\circ}+1 \end{array} \\ & \begin{array}{r} 2\left[\cos ^2 12^{\circ}+\sin ^2\left(30^{\circ}-12^{\circ}\right)\right. \left.+\sin ^2\left(30^{\circ}+12^{\circ}\right)\right] \end{array} \\ & +\begin{array}{r} {\left[\cos ^2 6^{\circ}+\sin ^2\left(30^{\circ}+6^{\circ}\right)+\sin ^2\left(30^{\circ}-6^{\circ}\right)\right]} +1 \end{array} \\ & \begin{array}{r} \because \cos ^2 \theta+\sin ^2\left(30^{\circ}+\theta\right)+\sin ^2\left(30^{\circ}-\theta\right)=0 \\ \cos ^2 \theta+\left[\sin 30^{\circ} \cos \theta+\cos 30^{\circ} \sin \theta\right]^2 \\ +\left[\sin 30^{\circ} \cos \theta-\cos 30^{\circ} \sin \theta\right]^2 \end{array} \\ & \text { Using }(a+b)^2+(a-b)^2=2\left(a^2+b^2\right) \end{aligned} $

$ \begin{aligned} & \begin{aligned} \Rightarrow & \cos ^2 \theta+2\left[\sin ^2 30^{\circ}\right. \cos ^2 \theta\left.\quad+\cos ^2 30^{\circ} \sin ^2 \theta\right] \end{aligned} \\ & \Rightarrow \cos ^2 \theta+2\left[\frac{1}{4} \cos ^2 \theta+\frac{3}{4} \sin ^2 \theta\right] \\ & \Rightarrow \cos ^2 \theta+\frac{1}{2} \cos ^2 \theta+\frac{3}{2} \sin ^2 \theta \\ & \Rightarrow 3 / 2 \quad\left[\because \cos ^2 \theta+\sin ^2 \theta=1\right]\\ & \end{aligned} $

$ \begin{aligned} &\text { Using this result in Eq. (i), we get }\\ &2\left(\frac{3}{2}\right)+\frac{3}{2}+1=\frac{11}{2} \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} & \frac{\sin A+\sin B+\sin C}{\sin ^2 \frac{A}{2}+\sin ^2 \frac{B}{2}+\sin ^2 \frac{C}{2}-1} \\ & =\frac{2 \sin \left(\frac{A+C}{2}\right) \cos \left(\frac{A-C}{2}\right)+2 \sin \frac{B}{2} \cos \frac{B}{2}}{\sin ^2 \frac{A}{2}-\cos ^2 \frac{C}{2}+\sin ^2 \frac{B}{2}} \\ & =\frac{2 \cos \frac{B}{2} \cos \frac{A-C}{2}+2 \sin \frac{B}{2} \cos \frac{B}{2}}{-\cos \left(\frac{A+C}{2}\right) \cos \left(\frac{A-C}{2}\right)+\sin ^2 \frac{B}{2}} \\ & {\left[\because \sin ^2 A-\cos ^2 B=-\cos (A+B) \cos (A-B)\right]} \\ & \quad=\frac{2 \cos \frac{B}{2}\left[\cos \left(\frac{A-C}{2}\right)+\sin \frac{B}{2}\right]}{-\sin \frac{B}{2}\left[\cos \left(\frac{A-C}{2}\right)+\sin \frac{B}{2}\right]} \\ & \quad=-2 \cot \frac{B}{2} \end{aligned} \end{aligned} $

We start with the given equations:

$ \begin{aligned} \cos \alpha + 4 \cos \beta + 9 \cos \gamma &= 0 \quad ...(i) \\ \sin \alpha + 4 \sin \beta + 9 \sin \gamma &= 0 \quad ...(ii) \end{aligned} $

Combine these equations using the complex form of the equations. Multiply both expressions:

$ \cos \theta + i \sin \theta = e^{i\theta} $

Thus, combining both equations, we have:

$ \begin{aligned} & (\cos \alpha + i \sin \alpha) + 4 (\cos \beta + i \sin \beta) + 9 (\cos \gamma + i \sin \gamma) = 0 \\ & e^{i\alpha} + 4 e^{i\beta} + 9 e^{i\gamma} = 0 \\ & \Rightarrow 1 + 4 e^{i(\beta-\alpha)} = -9 e^{i(\gamma-\alpha)} \end{aligned} $

Substitute using Euler's formula:

$ \begin{aligned} 1 + [4 \cos(\beta-\alpha) + 4i \sin(\beta-\alpha)] = -[9 \cos(\gamma-\alpha) + 9i \sin(\gamma-\alpha)] \end{aligned} $

Next, square both sides and compare the real parts:

$ \begin{aligned} & [1 + 4 \cos(\beta-\alpha)]^2 - 16 \sin^2(\beta-\alpha) \\ &= [9 \cos(\gamma-\alpha)]^2 - [9 \sin(\gamma-\alpha)]^2 \\ &= 1 + 8 \cos(\beta-\alpha) + 16[\cos^2(\beta-\alpha) - \sin^2(\beta-\alpha)] \\ &= 81[\cos^2(\gamma-\alpha) - \sin^2(\gamma-\alpha)] \end{aligned} $

Use the identities for double angles:

$ \begin{aligned} & \Rightarrow 81 \cos 2(\gamma-\alpha) - 16 \cos 2(\beta-\alpha) \\ &= 1 + 8 \cos(\beta-\alpha) \end{aligned} $

Therefore, the expression $81 \cos (2\gamma - 2\alpha) - 16 \cos (2\beta - 2\alpha)$ simplifies to:

$ 1 + 8 \cos (\beta-\alpha) $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} & \tan \alpha+2 \tan 2 \alpha+4 \tan 4 \alpha+8 \cot 8 \alpha \\ & =\tan \alpha+2 \tan 2 \alpha+4 \tan 4 \alpha +8 \times \frac{1-\tan ^2 4 \alpha}{2 \tan 4 \alpha} \\ & =\tan \alpha+2 \tan 2 \alpha +\frac{8 \tan ^2 4 \alpha+8-8 \tan ^2 4 \alpha}{2 \tan 4 \alpha} \\ & =\tan \alpha+2 \tan 2 \alpha+\frac{8}{2\left(\frac{2 \tan 2 \alpha}{1-\tan ^2 2 \alpha}\right)} \\ & =\tan \alpha+2 \tan 2 \alpha+\frac{4\left(1-\tan ^2 2 \alpha\right)}{2 \tan 2 \alpha} \\ & =\tan \alpha+\frac{4 \tan ^2 2 \alpha+4-4 \tan ^2 2 \alpha}{2 \tan 2 \alpha} \\ & =\tan \alpha+\frac{2}{\tan 2 \alpha}=\tan \alpha+\frac{2\left(1-\tan ^2 \alpha\right)}{2 \tan \alpha} \\ & =\frac{2 \tan ^2 \alpha+2-2 \tan ^2 \alpha}{2 \tan \alpha}=\frac{1}{\tan \alpha}=\cot \alpha \end{aligned} \end{aligned} $

$ \text { } \begin{aligned} \tan 9^{\circ}-\tan 27^{\circ}-\tan 63^{\circ}+\tan 81^{\circ} \\ =\left(\tan 9^{\circ}+\tan 81^{\circ}\right)-\left(\tan 27^{\circ}+\tan 63^{\circ}\right) \\ =\left(\tan 9^{\circ}+\cot 9^{\circ}\right)-\left(\tan 27^{\circ}+\cot 27^{\circ}\right) \\ =\left(\frac{\cos 9^{\circ}}{\sin 9^{\circ}}+\frac{\sin 9^{\circ}}{\cos 9^{\circ}}\right)-\left(\frac{\sin 27^{\circ}}{\cos 27^{\circ}}+\frac{\cos 27^{\circ}}{\sin 27^{\circ}}\right) \end{aligned} $

$ \begin{aligned} & =\frac{\cos ^2 9^{\circ}+\sin ^2 9^{\circ}}{\sin 9^{\circ} \cdot \cos 9^{\circ}}-\frac{\sin ^2 27^{\circ}+\cos ^2 27^{\circ}}{\cos 27^{\circ} \cdot \sin 27^{\circ}} \\ & =\frac{1}{\sin 9^{\circ} \cdot \cos 9^{\circ}}-\frac{1}{\cos 27^{\circ} \cdot \sin 27^{\circ}} \\ & =\frac{2}{2 \sin 9^{\circ} \cos 9^{\circ}}-\frac{2}{2 \cos 27^{\circ} \sin 27^{\circ}} \\ & =\frac{2}{\sin 18^{\circ}}-\frac{2}{\sin 54^{\circ}}=\left(\frac{\sin 54^{\circ}-\sin 18^{\circ}}{\sin 18^{\circ} \cdot \sin 54^{\circ}}\right) \\ & =2 \cdot \frac{2 \cos 36^{\circ} \cdot \sin 18^{\circ}}{\cos 36^{\circ} \cdot \sin 18^{\circ}}=4 \end{aligned} $

$ \text { } \begin{aligned} & \cos 6^{\circ} \cdot \sin 24^{\circ} \cos 72^{\circ} \\ &=\frac{1}{2} \cdot 2 \cos 6^{\circ} \sin 24^{\circ} \cos 72^{\circ} \\ &=\frac{1}{2}\left(\sin 30^{\circ}+\sin 18^{\circ}\right) \cos 72^{\circ} \\ &=\frac{1}{2}\left[\left(\frac{1}{2}+\frac{\sqrt{5}-1}{4}\right)\left(\frac{\sqrt{5}-1}{4}\right)\right]=\frac{1}{8} \end{aligned} $

Given, $\sin h(x)=\frac{\sqrt{21}}{2}$

$ \begin{aligned} & \Rightarrow \cosh x=\sqrt{1+\sinh ^2 x}=\sqrt{1+\frac{21}{4}} \\ & \Rightarrow \cosh x=\frac{5}{2} \end{aligned} $

Now, $\cosh 2 x+\sinh 2 x=1+2 \sinh ^2 x+2 \sinh x \cosh$

$ \begin{aligned} & =1+2 \cdot \frac{21}{4}+2 \times \frac{\sqrt{21}}{2} \times \frac{5}{2} \\ & =\frac{23}{2}+\sqrt{21} \times \frac{5}{2}=\frac{23+5 \sqrt{21}}{2} \end{aligned} $

$\frac{1}{11 \cos 2 x+60 \sin 2 x+69}$

denominator $11 \cos 2 x+60 \sin 2 x+69$

To find extrema of denominator

$11 \cos 2 x+60 \sin 2 x=\sqrt{11^2+60^2} \sin (2 x+\theta)$

where, $\theta=\tan ^{-1} \frac{11}{60}$

$=61 \sin (2 x+\theta)$

This minimum value is -61 and the maximum value is 61 .

Therefore, $11 \cos 2 x+60 \sin 2 x+69$

$ =69-61=8 \text { to } 69+61=130 $

The maximum value is 130 and minimum is 8 of denominator

Hence, $M_1=\frac{1}{8}$

For, $3 \cos ^2 5 x+4 \sin ^2 5 x$

$ \begin{aligned} & =3\left(1-\sin ^2 5 x\right)+4 \sin ^2 5 x \\ & =3+\sin ^2 5 x \end{aligned} $

The maximum value of $\sin ^2 5 x$ is 1 , hence

$ 3+\sin ^2 5 x=3+1=4 $

because the range of (sine) ${ }^2$ is between 0 and 1

The minimum value of $\sin ^2 5 x$ is 0 .

Hence, $3+\sin ^2 5 x \geq 3+0=3$

Thus, the maximum value $M_2=4$

Finally the ratio is

$ \frac{M_1}{M_2}=\frac{\frac{1}{8}}{4}=\frac{1}{8 \times 4} \Rightarrow \frac{M_1}{M_2}=\frac{1}{32} $

To solve the expression:

$ 4 \cos \frac{\pi}{7} \cos \frac{\pi}{5} \cos \frac{2 \pi}{7} \cos \frac{2 \pi}{5} \cos \frac{4 \pi}{7} $

we can use known trigonometric identities and values. Direct computation might be cumbersome, so we leverage a known result for products of cosines.

Here is the important formula:

$ \cos \left(\frac{\pi}{7}\right) \cos \left(\frac{2\pi}{7}\right) \cos \left(\frac{3\pi}{7}\right) = \frac{1}{8} $

Given the symmetry properties of cosine, specifically that:

$ \cos \left(\frac{4\pi}{7}\right) = -\cos \left(\frac{3\pi}{7}\right), $

we can deduce:

$ \cos \frac{\pi}{7} \cos \frac{2\pi}{7} \cos \frac{4\pi}{7} = \frac{-1}{8} \tag{i} $

Additionally, using the identity for the product of cosines:

$ \cos \frac{\pi}{5} \cos \frac{2\pi}{5} = \frac{1}{4} \tag{ii} $

Combining these results from equations (i) and (ii):

$ \begin{align*} 4 \cos \frac{\pi}{7} \cos \frac{\pi}{5} \cos \frac{2\pi}{7} \cos \frac{2\pi}{5} \cos \frac{4\pi}{7} &= 4 \times \left(-\frac{1}{8}\right) \times \frac{1}{4} \\ &= \frac{-1}{8} \end{align*} $

Hence, the value of the expression is $-\frac{1}{8}$.

Step 1: Solve for $\tanh x = \frac{3}{5}$:

The hyperbolic tangent function is defined as:

$ \tanh x = \frac{\sinh x}{\cosh x} = \frac{3}{5} $

This implies:

$ \sinh x = \frac{3}{5} \cosh x $

Let $\cosh x = z$, then:

$ \sinh x = \frac{3}{5}z $

Using the identity:

$ \cosh^2 x - \sinh^2 x = 1 $

Substitute the values:

$ z^2 - \left(\frac{3}{5}z\right)^2 = 1 $

Solve this equation:

$ z^2 - \frac{9}{25}z^2 = 1 $

$ \frac{16}{25}z^2 = 1 \quad \Rightarrow \quad z^2 = \frac{25}{16} $

Thus,

$ z = \frac{5}{4} \quad \Rightarrow \quad \cosh x = \frac{5}{4}, \, \sinh x = \frac{3}{4} $

Step 2: Solve for $\operatorname{sech} y = \frac{3}{5}$:

The hyperbolic secant is defined as:

$ \operatorname{sech} y = \frac{1}{\cosh y} = \frac{3}{5} $

Therefore:

$ \cosh y = \frac{5}{3} $

Using the identity for hyperbolic functions:

$ \cosh^2 y - \sinh^2 y = 1 $

Calculate $\sinh y$:

$ \left(\frac{5}{3}\right)^2 - 1 = \sinh^2 y $

$ \sinh^2 y = \frac{16}{9} \quad \Rightarrow \quad \sinh y = \frac{4}{3} $

Step 3: Calculate $e^{x+y}$:

We need to find $e^{x} \cdot e^{y}$.

For $e^x$:

$ e^x = \cosh x + \sinh x = \frac{5}{4} + \frac{3}{4} = 2 $

For $e^y$:

$ e^y = \cosh y + \sinh y = \frac{5}{3} + \frac{4}{3} = 3 $

Thus:

$ e^{x+y} = e^x \cdot e^y = 2 \cdot 3 = 6 $

Therefore, the value of $e^{x+y}$ is 6.

Given that $A, B$ and $C$ are the angle of triangle

$ \begin{aligned} A+B+C & =\pi \\ 2 A+2 B+2 C & =2 \pi \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Now, $\sin 2 A-\sin 2 B+\sin 2 C$

$ \begin{aligned} &=\sin 2 A+\sin 2 C-\sin 2 B \\ &= 2 \sin \left(\frac{2 A+2 C}{2}\right) \cos \left(\frac{2 A-2 C}{2}\right)-\sin [2 \pi-2(A+C)] \end{aligned} $

[from Eq. (i)]

$ \begin{aligned} & =2 \sin (A+C) \cos (A-C)+\sin 2(A+C) \\ & =2 \sin (A+C) \cos (A-C)+2 \sin (A+C) \cos (A+C) \\ & =2 \sin (A+C)[\cos (A-C)+\cos (A+C)] \\ & =2 \sin (\pi-B)[2 \cos A \cos C] \\ & =4 \cos A \sin B \cos C \end{aligned} $

In case of assertion, if $A=10^{\circ}$,

$ \begin{aligned} & B=16^{\circ} \text { and } C=19^{\circ} \\ & \because \quad A+B+C=180^{\circ} \\ & \Rightarrow \quad 2 A+2 B+2 C=\frac{\pi}{2} \\ & \Rightarrow \quad 2 A+2 B=\frac{\pi}{2}-2 C \end{aligned} $

On taking tan both sides, we get

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

So, assertion is true.

In case of reason,

$ \text { Since, } A+B+C=180^{\circ} $

$ \begin{aligned} & \frac{A}{2}+\frac{B}{2}+\frac{C}{2}=\frac{\pi}{2} \\ \Rightarrow \quad & \frac{A}{2}+\frac{B}{2}=\frac{\pi}{2}-\frac{C}{2} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \Rightarrow \cot \left(\frac{A}{2}+\frac{B}{2}\right)=\cot \left(\frac{\pi}{2}-\frac{C}{2}\right) \\ & \Rightarrow \frac{\cot \frac{A}{2} \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}+\cot \frac{B}{2}+\cot \frac{C}{2} \end{aligned}\\ &\text { So, reason is true. } \end{aligned} $

$ \text { Given, } \tan \alpha=\frac{1}{7}, \sin \beta=\frac{1}{\sqrt{10}} $

$ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\tan \,\alpha=\frac{p}{b}=\frac{1}{7} $

Using Pythagoras theorem,

$ \begin{aligned} & \quad P^2+b^2=a^2 \Rightarrow a=\sqrt{1^2+7^2}=\sqrt{50} \\ & \sin \alpha=\frac{1}{\sqrt{50}} \text { and } \cos \alpha=\frac{7}{\sqrt{50}} \\ & \text { and } \sin \beta=\frac{1}{\sqrt{10}} \\ & \cos ^2 \beta=1-\left(\frac{1}{\sqrt{10}}\right)^2 \quad\left[\because \sin ^2 \alpha+\cos ^2 \alpha=1\right] \\ & \Rightarrow \cos ^2 \beta=\frac{9}{10} \Rightarrow \cos \beta=-\frac{3}{\sqrt{10}} \end{aligned} $

[ $\because \beta$ is in 3rd quadrant $]$

$ \text { Now, } \sin (2 \alpha+\beta)=\sin 2 \alpha \cos \beta +\cos 2 \alpha \sin \beta$

$ \sin 2 \alpha=2 \sin \alpha \cos \alpha=2 \cdot \frac{1}{\sqrt{50}} \cdot \frac{7}{\sqrt{50}}=\frac{14}{50} $

$ \cos 2 \alpha=\cos ^2 \alpha-\sin ^2 \alpha=\frac{49}{50}-\frac{1}{50}=\frac{48}{50} $

On putting all these values in Eq. (i), we gac

$ \begin{aligned} & \sin (2 \alpha+\beta)=\frac{14}{50} \cdot\left(-\frac{3}{\sqrt{10}}\right)+\frac{48}{50} \cdot \frac{1}{\sqrt{10}} \\ & =-\frac{42}{50 \sqrt{10}}+\frac{48}{50 \sqrt{10}}=\frac{1}{50 \sqrt{10}}(48-42) \end{aligned} $

$ =\frac{3}{25 \sqrt{10}} $

Let $f(x)=\frac{\tan (5 x) \cos 3 x}{\sin 6 x}$

1.Period of $\tan (5 x)=\frac{\pi}{5}$

2.Period of $\cos 3 x=\frac{2 \pi}{3}$

3.Period of $\sin 6 x=\frac{\pi}{3}$

The period of $f(x)$ is the least common multiple of $\frac{\pi}{5}, \frac{2 \pi}{3}$ and $\frac{\pi}{3}$.

$ \operatorname{LCM}\left(\frac{\pi}{3}, \frac{2 \pi}{3}, \frac{\pi}{3}\right)=2 \pi $

$ \therefore \alpha=2 \pi $

If $\sin x+\sin y=\alpha$

$ \cos x+\cos y=\beta $

$\operatorname{cosec}(x+y)=$ ?

From Eq. (i), we get

$ \begin{aligned} & 2 \sin \left(\frac{x+y}{2}\right) \cdot \cos \left(\frac{x-y}{2}\right)=\alpha \\ & \left(\because \sin C+\sin D=2 \sin \frac{(C+D)}{2} \cos \frac{(C-D)}{2}\right) \end{aligned} $

and using Eq. (ii)

$ \begin{aligned} & 2 \cos \left(\frac{x-y}{2}\right) \cdot \cos \left(\frac{x+y}{2}\right)=\beta \\ & \left(\because \cos C+\cos D=2 \cos \left(\frac{C-D}{2}\right) \cdot \cos \left(\frac{C+D}{2}\right)\right) \end{aligned} $

On dividing Eq. (iii) by Eq. (iv), we get

$ \begin{aligned} & \tan \left(\frac{x+y}{2}\right)=\frac{\alpha}{\beta} \\ & \left(\because \tan ^2 \theta=\sec ^2 \theta-1 \text { and } \sec \theta=\frac{1}{\cos \theta}\right) \end{aligned} $

So, $\cos \left(\frac{x+y}{2}\right)=\left[\frac{\beta^2}{\alpha^2+\beta^2}\right]^{\frac{1}{2}}$

$ \begin{aligned} \sin \left(\frac{x+y}{2}\right) & =\left(1-\cos ^2\left(\frac{x+y}{2}\right)\right)^{\frac{1}{2}} \\ & =\left[\frac{\alpha^2}{\alpha^2+\beta^2}\right]^{\frac{1}{2}} \end{aligned} $

$ \begin{array}{r} \sin (x+y)=2 \sin \left(\frac{x+y}{2}\right) \cdot \cos \left(\frac{x+y}{2}\right) \\ \quad\left(\because \sin A=\frac{2 \sin A}{2} \cos \frac{A}{2}\right) \end{array} $

From Eqs. (vi) and (vii), we get

$ \begin{aligned} \sin (x+y) & =\frac{2 \alpha}{\left(\alpha^2+\beta^2\right)^{\frac{1}{2}}} \cdot \frac{\beta}{\left(\alpha^2+\beta^2\right)^{\frac{1}{2}}} \\ & =\frac{2 \alpha \beta}{\left(\alpha^2+\beta^2\right)}\left(\because \operatorname{cosec} \theta=\frac{1}{\sin \theta}\right) \\ \operatorname{cosec}(x+y) & =\frac{1}{\sin (x+y)} \\ & =\frac{1}{\frac{2 \alpha \beta}{\left(\alpha^2+\beta^2\right)}}=\frac{\alpha^2+\beta^2}{2 \alpha \beta} \end{aligned} $

Given, $P+Q+R=\frac{\pi}{4}$

$ \therefore \frac{P}{2}+\frac{Q}{2}+\frac{R}{2}=\frac{\pi}{8} $

Now, $\cos \left(\frac{\pi}{8}-P\right)+\cos \left(\frac{\pi}{8}-Q\right)+\cos \left(\frac{\pi}{8}-R\right)$

$ \begin{aligned} & =2 \cos \left(\frac{\left(\frac{\pi}{8}-P\right)+\left(\frac{\pi}{8}-Q\right)}{2}\right) \\ & \cos \left(\frac{\left(\frac{\pi}{8}-P\right)-\left(\frac{\pi}{8}-Q\right)}{2}\right)+\cos \left(\frac{\pi}{8}-R\right) \\ & =2 \cos \left(\frac{\pi}{8}-\frac{P}{2}-\frac{Q}{2}\right) \cos \left(\frac{Q}{2}-\frac{P}{2}\right)+\cos \left(\frac{\pi}{8}-R\right) \\ & =2 \cos \left(\frac{R}{2}\right) \cos \left(\frac{Q}{2}-\frac{P}{2}\right)+\cos \left(\frac{P}{2}+\frac{Q}{2}-\frac{R}{2}\right) \\ & =2 \cos \left(\frac{R}{2}\right) \cos \left(\frac{Q}{2}-\frac{P}{2}\right)+\cos \left(\frac{P}{2}+\frac{Q}{2}\right) \\ & =2 \cos \left(\frac{R}{2}\right)+\sin \left(\frac{P}{2}+\frac{Q}{2}\right) \sin \left(\frac{R}{2}\right) \\ & \left.-\cos \left(\frac{Q}{2}-\frac{P}{2}\right)+\cos \left(\frac{P}{2}+\frac{Q}{2}\right)\right] \\ & =2 \cos \left(\frac{R}{2}+\frac{Q}{2}\right) \cos \left(\frac{R}{2}\right)+\sin \left(\frac{P}{2}+\frac{Q}{2}\right) \sin \left(\frac{R}{2}\right) \\ & -\left[\cos \left(\frac{P}{2}+\frac{P}{2}+\frac{Q}{2}\right) \cos \left(\frac{R}{2}\right)-\cos \left(\frac{Q}{2}-\frac{P}{2}\right)\right] \end{aligned} $

$ \begin{aligned} & =2 \cos \left(\frac{R}{2}\right)\left[2 \cos \left(\frac{Q}{2}\right) \cos \left(\frac{P}{2}\right)\right] -\cos \left(\frac{P}{2}+\frac{Q}{2}+\frac{R}{2}\right) \\ & =4 \cos \left(\frac{P}{2}\right) \cos \left(\frac{Q}{2}\right) \cos \left(\frac{R}{2}\right)-\cos \left(\frac{\pi}{8}\right) \end{aligned} $

Given that $\theta$ is an acute angle, $\cosh x = k$, and $\sinh x = \tan \theta$, we can establish the following relationships:

First, note that:

$ \tan \theta = \sinh x $

Squaring both sides, we obtain:

$ \tan^2 \theta = \sinh^2 x $

We also have the trigonometric identity:

$ \sec^2 \theta = 1 + \tan^2 \theta $

Substituting $\sinh^2 x$ for $\tan^2 \theta$ gives:

$ \sec^2 \theta = 1 + \sinh^2 x $

Next, recall the identity for hyperbolic functions:

$ \cosh^2 x - \sinh^2 x = 1 $

This implies:

$ \cosh^2 x = 1 + \sinh^2 x $

Given $\cosh x = k$, it follows that

$ \cosh^2 x = k^2 $

Thus:

$ \sec^2 \theta = \cosh^2 x = k^2 $

Taking the reciprocal to find $\cos^2 \theta$:

$ \cos^2 \theta = \frac{1}{k^2} $

Using the Pythagorean identity:

$ 1 = \sin^2 \theta + \cos^2 \theta $

Substitute the expression for $\cos^2 \theta$:

$ 1 = \sin^2 \theta + \frac{1}{k^2} $

Rearranging for $\sin^2 \theta$ gives:

$ \sin^2 \theta = 1 - \frac{1}{k^2} $

This simplifies to:

$ \sin^2 \theta = \frac{k^2 - 1}{k^2} $

Taking the square root, given that $\theta$ is an acute angle (which ensures $\sin \theta$ is positive), we find:

$ \sin \theta = \frac{\sqrt{k^2 - 1}}{k} $

Given:

$ \sec \theta + \tan \theta = \frac{1}{3} \quad \ldots \text {(i)} $

We know the identity:

$ \sec^2 \theta - \tan^2 \theta = 1 $

This can be rewritten as:

$ (\sec \theta + \tan \theta)(\sec \theta - \tan \theta) = 1 $

Substituting the given value from equation (i):

$ \frac{1}{3} (\sec \theta - \tan \theta) = 1 $

Solving for $\sec \theta - \tan \theta$:

$ \sec \theta - \tan \theta = 3 \quad \ldots \text {(ii)} $

Using equations (i) and (ii), we solve for $\sec \theta$ and $\tan \theta$:

$ \begin{aligned} \sec \theta &= \frac{5}{3} \\ \tan \theta &= \frac{-4}{3} \end{aligned} $

Now, we find $\sin \theta$:

$ \sin \theta = \frac{\tan \theta}{\sec \theta} = \frac{\frac{-4}{3}}{\frac{5}{3}} = \frac{-4}{5} $

Here, both $\sin \theta$ and $\tan \theta$ are negative, indicating that $\theta$ lies in the IVth quadrant.

Calculating $\sin 2\theta$:

$ \sin 2\theta = 2 \sin \theta \cos \theta = 2 \times \frac{-4}{5} \times \frac{3}{5} = \frac{-24}{25} $

Next, calculating $\tan 2\theta$:

$ \tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta} = \frac{24}{7} $

Since $\sin 2\theta$ is negative and $\tan 2\theta$ is positive, $2\theta$ lies in the IIIrd quadrant.

Since, $ 540^{\circ}<{A}<630^{\circ} $

$ \therefore \quad 270^{\circ}<\frac{A}{2}<315^{\circ} $

Since, $A$ lies in IIIrd quadrant $\therefore \tan A$ is positive and $\frac{A}{2}$ lies in IVth quadrant $\therefore \tan \frac{A}{2}$ is negative.

Now, $\cos A=\frac{5}{13}$

$\Rightarrow \quad \tan A=\frac{12}{5}$

Also, $\tan A=\frac{2 \tan \frac{A}{2}}{1-\tan ^2 \frac{A}{2}}$

$\Rightarrow \quad \frac{12}{5}-\frac{12}{5} \tan ^2 \frac{A}{2}=2 \tan \frac{A}{2}$

$\Rightarrow \quad 6 \tan ^2 \frac{A}{2}+5 \tan \frac{A}{2}-6=0$

$\Rightarrow \quad \tan \frac{A}{2}=\frac{-3}{2}$

$\therefore \quad \tan \frac{A}{2} \tan A=\frac{-18}{5}$

Given, $3 \sin (\alpha-\beta)=5 \cos (\alpha+\beta)$

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

Divide both numerator and denominator by $\cos \alpha \cos \beta$

$ \begin{aligned} & \frac{\tan \alpha-\tan \beta}{1-\tan \alpha \tan \beta}=\frac{5}{3} \\ \Rightarrow & 3 \tan \alpha-3 \tan \beta=5-5 \tan \alpha \tan \beta \\ \Rightarrow & 3 \tan \alpha-3 \tan \beta-5+5 \tan \alpha \tan \beta=0 \end{aligned} $

Now, $\tan \left(\frac{\pi}{4}+\alpha\right)+4 \tan \left(\frac{\pi}{4}+\beta\right)$

$ \begin{aligned} & =\frac{1+\tan \alpha}{1-\tan \alpha}+4\left(\frac{1+\tan \beta}{1-\tan \beta}\right) \\ & =\frac{(1-\tan \beta+\tan \alpha-\tan \alpha \tan \beta+4}{(1-\tan \alpha)(1-\tan \beta)} \\ & =\frac{3 \tan \beta-3 \tan \alpha+5-5 \tan \alpha \tan \beta}{(1-\tan \alpha)(1-\tan \beta)} \\ & =\frac{-(3 \tan \alpha-3 \tan \beta-5+5 \tan \alpha \tan \beta)}{(1-\tan \alpha)(1-\tan \beta)} \\ & =0 \end{aligned} $

We have,

$\cos \alpha+\cos \beta+\cos \gamma=\sin \alpha+\sin \beta+\sin \gamma=0$ then,

$ \begin{aligned} & \left(\cos ^3 \alpha+\cos ^3 \beta+\cos ^3 \gamma\right)^2+ \\ & \quad\left(\sin ^3 \alpha+\sin ^3 \beta+\sin ^3 \gamma\right)^2 \\ & =(3 \cos \alpha \cos \beta \cos \gamma)^2+(3 \sin \alpha \sin \beta \sin \gamma)^2 \\ & {\left[\begin{array}{c} \because \text { If } x+y+z=0 \\ \therefore x^3+y^3+z^3=3 x y z \end{array}\right]} \\ & =9(\cos \alpha \cos \beta \cos \gamma)^2+9(\sin \alpha \sin \beta \sin \gamma)^2 \end{aligned} $

So, let $\alpha=\theta, \beta=\theta+\frac{2 \pi}{3}$ and $\gamma=\theta+\frac{4 \pi}{3}$

then, $\cos \alpha \cos \beta \cos \gamma=\cos \theta \cos \left(\theta+\frac{2 \pi}{3}\right)$

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

Similarly,

$ \sin \alpha \sin \beta \sin \gamma=\frac{3}{4} \sin \theta \cos ^2 \theta-\frac{1}{4} \sin ^3 \theta $

From Eqs. (ii) and (iii) in Eq. (i), we get

$ \begin{aligned} & \left(\cos ^3 \alpha+\cos ^2 \beta+\cos ^3 \gamma\right)^2 \\ & \quad+\left(\sin ^3 \alpha+\sin ^3 \beta+\sin ^3 \gamma\right)^2 \end{aligned} $

$ \begin{aligned} & =\frac{9\left(\frac{1}{4} \cos ^3 \theta-\frac{3}{4} \cos \theta \sin ^2 \theta\right)^2+9}{} \\ & \quad\left(\frac{3}{4} \sin \theta \cos ^2 \theta-\frac{1}{4} \sin ^3 \theta\right)^2 \\ & =\frac{9}{16}\left[\left(\cos ^3 \theta-3 \cos \theta \sin ^2 \theta\right)^2\right. \\ & \\ & \left.\quad+\left(3 \sin \theta \cos ^2 \theta-\sin ^3 \theta\right)^2\right] \\ & =\frac{9}{16}\left[\cos ^6 \theta+9 \cos ^2 \theta \sin ^4 \theta\right. \\ & \quad-6 \cos ^4 \theta \sin ^2 \theta+9 \sin ^2 \theta \cos ^4 \theta \\ & =\frac{9}{16}\left[\cos ^6 \theta+\sin ^6 \theta-6 \cos ^2 \theta \sin ^4 \theta\right] \\ & =\frac{9}{16}\left[\left(\cos ^2 \theta\right)^3 \theta \sin ^4 \theta\right. \\ & =\left(\sin ^2 \theta\right)^3+3 \cos ^2 \theta \\ & =\frac{9}{16}\left(\cos ^2 \theta+\sin ^2 \theta\right)^3 \\ & =\frac{9}{16} \times(1)^3=\frac{9}{16} \end{aligned} $