Trigonometric Ratio and Identites

$ \begin{aligned} & \tan 2 \alpha=\tan [(\alpha+\beta)+(\alpha-\beta)] \\ & \tan 2 \alpha=\frac{\tan (\alpha+\beta)+\tan (\alpha-\beta)}{1-\tan (\alpha+\beta) \cdot \tan (\alpha-\beta)} \\ & \tan 2 \alpha=\frac{\left(-\sqrt{99}+\frac{3}{\sqrt{55}}\right)}{1-(\sqrt{99})\left(\frac{3}{\sqrt{55}}\right)} \\ & \tan 2 \alpha=\frac{-3 \sqrt{11}+\frac{3}{\sqrt{5} \times \sqrt{11}}}{1+\frac{9 \sqrt{11}}{\sqrt{5} \times \sqrt{11}}} \\ & \tan 2 \alpha=\frac{3(1-11 \sqrt{5})}{\sqrt{11}(9+\sqrt{5})} \\ & r=11, s=9 \\ & r+s=20 \end{aligned} $

$ \begin{aligned} & \frac{\cos 60^0 \cos 36^0}{\sin 30^0 \cdot \sin 18^0}=\frac{\sqrt{5}+1}{4} \cdot \frac{4}{\sqrt{5}-1}=\frac{3+\sqrt{5}}{2} \\ & \alpha+\beta=4 \end{aligned} $

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

$\begin{aligned} & \tan 9^{\circ}-\tan 27^{\circ}-\tan 63^{\circ}+\tan 81^{\circ} \\\\ & =\tan 9^{\circ}+\tan \left(90^{\circ}-9^{\circ}\right)-\tan 27^{\circ}-\tan \left(90^{\circ}-27^{\circ}\right) \\\\ & =\tan 9^{\circ}+\cot 9^{\circ}-\tan 27^{\circ}-\cot 27^{\circ} \\\\ & =\frac{\sin 9^{\circ}}{\cos 9^{\circ}}+\frac{\cos 9^{\circ}}{\sin 9^{\circ}}-\left(\frac{\sin 27^{\circ}}{\cos 27^{\circ}}+\frac{\cos 27^{\circ}}{\sin 27^{\circ}}\right) \\\\ & =\frac{\sin ^2 9^{\circ}+\cos ^2 9^{\circ}}{\sin 9^{\circ} \cos 9^{\circ}}-\left(\frac{\sin ^2 27^{\circ}+\cos ^2 27^{\circ}}{\cos 27^{\circ} \sin 27^{\circ}}\right) \\\\ & =\frac{2}{\sin 18^{\circ}}-\frac{2}{\sin 54^{\circ}} \\\\ & =\frac{2 \times 4}{\sqrt{5}-1}-\frac{2 \times 4}{\sqrt{5}+1} \\\\ & =8\left(\frac{\sqrt{5}+1-\sqrt{5}+1}{5-1}\right)=2(2)=4\end{aligned}$

$(\sin 10^\circ \,.\,\sin 50^\circ \,.\,\sin 70^\circ )\,.\,(\sin 10^\circ \,.\,\sin 20^\circ \,.\,\sin 40^\circ )$

$ = \left( {{1 \over 4}\sin 30^\circ } \right)\,.\,\left[ {{1 \over 2}\sin 10^\circ (\cos 20^\circ - \cos 60^\circ )} \right]$

$ = {1 \over {16}}\left[ {\sin 10^\circ \left( {\cos 20^\circ - {1 \over 2}} \right)} \right]$

$ = {1 \over {32}}[2\sin 10^\circ \,.\,\cos 20^\circ - \sin 10^\circ ]$

$ = {1 \over {32}}[\sin 30^\circ - \sin 10^\circ - \sin 10^\circ ]$

$ = {1 \over {64}} - {1 \over {64}}\sin 10^\circ $

Clearly, $\alpha = {1 \over {64}}$

Hence $16 + {\alpha ^{ - 1}} = 80$

We know,

$ - \sqrt {{a^2} + {b^2}} \le a\cos x + b\sin x \le \sqrt {{a^2} + {b^2}} $

$ \therefore $ $ - \sqrt {{3^2} + {4^2}} \le 3\cos x + 4\sin x \le \sqrt {{3^2} + {4^2}} $

$ - 5 \le k + 1 \le 5$

$ - 6 \le k \le 4$

$ \therefore $ Set of integers = $ - 6, - 5, - 4, - 3, - 2, - 1,0,1,2,3,4$ = Total 11 intergers.

${{\sqrt 2 \sin \alpha } \over {\sqrt {1 + \cos 2\alpha } }} = {1 \over 7}$

$ \Rightarrow $ ${{\sqrt 2 \sin \alpha } \over {\sqrt {2{{\cos }^2}\alpha } }}$ = ${1 \over 7}$

$ \Rightarrow $ ${{\sqrt 2 \sin \alpha } \over {\sqrt 2 \cos \alpha }}$ = ${1 \over 7}$

$ \Rightarrow $ tan$\alpha $ = ${1 \over 7}$

Also given $\sqrt {{{1 - \cos 2\beta } \over 2}} = {1 \over {\sqrt {10} }}$

$ \Rightarrow $ ${{\sqrt 2 \sin \beta } \over {\sqrt 2 }}$ = ${1 \over {\sqrt {10} }}$

$ \Rightarrow $ sin $\beta $ = ${1 \over {\sqrt {10} }}$

$ \therefore $ tan $\beta $ = ${1 \over 3}$

$\tan 2\beta = {{2\tan \beta } \over {1 - {{\tan }^2}\beta }}$

= ${{2\left( {{1 \over 3}} \right)} \over {1 - {1 \over 9}}}$ = ${3 \over 4}$

$ \therefore $ tan($\alpha $ + 2$\beta $) = ${{\tan \alpha + \tan 2\beta } \over {1 - \tan \alpha .\tan 2\beta }}$

= ${{{1 \over 7} + {3 \over 4}} \over {1 - {1 \over 7}.{3 \over 4}}}$

= ${{25} \over {25}}$ = 1