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} $

$\begin{aligned} & \text { Consider } \mathrm{E}=\frac{\sin \theta}{\cos 3 \theta} \Rightarrow \mathrm{E}=\frac{2 \sin \theta \cos \theta}{2 \cos 3 \theta \cos \theta} \\ & \Rightarrow \mathrm{E}=\frac{\sin 2 \theta}{2 \cos 3 \theta \cos \theta} \Rightarrow \mathrm{E}=\frac{\sin (30-\theta)}{2 \cos 3 \theta \cos \theta} \\ & \Rightarrow \mathrm{E}=\frac{1}{2}[\tan 3 \theta-\tan \theta] \\ & \mathrm{A}=\frac{1}{2}\left[\tan 9^{\circ}-\tan 3^{\circ}+\tan 27^{\circ}-\tan 9^{\circ}+\tan 81^{\circ}-\tan 27^{\circ}\right] \\ & \therefore \mathrm{A}=\frac{1}{2}\left[\tan 81^{\circ}-\tan 3^{\circ}\right] \\ & \therefore \frac{\mathrm{B}}{\mathrm{A}}=2\end{aligned}$

$ \vec a_k=(\tan\theta_k)\hat i+\hat j,\qquad \vec b_k=\hat i-(\cot\theta_k)\hat j, $

where

$ \theta_k=\frac{2^{k-1}\pi}{2^n+1},\qquad k=1,2,\dots,n. $

We have to find

$ \frac{\sum\limits_{k=1}^n |\vec a_k|^2}{\sum\limits_{k=1}^n |\vec b_k|^2}. $

Now let us compute the magnitudes.

For $\vec a_k$,

$ |\vec a_k|^2=(\tan\theta_k)^2+1=\sec^2\theta_k. $

For $\vec b_k$,

$ |\vec b_k|^2=1+(\cot\theta_k)^2=\csc^2\theta_k. $

So the required ratio becomes

$ \frac{\sum\limits_{k=1}^n \sec^2\theta_k}{\sum\limits_{k=1}^n \csc^2\theta_k}. $

Now use

$ \sec^2\theta=\frac{1}{\cos^2\theta},\qquad \csc^2\theta=\frac{1}{\sin^2\theta}. $

Thus,

$ \frac{\sec^2\theta_k}{\csc^2\theta_k} =\frac{1/\cos^2\theta_k}{1/\sin^2\theta_k} =\tan^2\theta_k, $

but this does not directly simplify the sum. So we look for a relation among the angles.

Given

$ \theta_k=\frac{2^{k-1}\pi}{2^n+1}, $

notice that

$ 2^n\pi=(2^n+1)\pi-\pi. $

Hence

$ \theta_{n+1-k} =\frac{2^{\,n-k}\pi}{2^n+1}. $

Now observe

$ \theta_k+\theta_{n+1-k} =\frac{2^{k-1}+2^{n-k}}{2^n+1}\pi. $

Instead of pairing this way, the key identity comes from the standard result:

If

$ \alpha_k=\frac{2^{k-1}\pi}{2^n+1}, $

then the set

$ \{2\alpha_1,2\alpha_2,\dots,2\alpha_n\} $

behaves cyclically modulo $\pi$, and one gets

$ \prod_{k=1}^n \tan\theta_k=\tan\theta_1\tan\theta_2\cdots\tan\theta_n=\sqrt{2^n+1}\text{ type identities,} $

but here we need a sum identity.

A much simpler way is to use the doubling-angle relation.

Let

$ x_k=\tan\theta_k. $

Since

$ \theta_{k+1}=2\theta_k, $

we have

$ x_{k+1}=\tan 2\theta_k=\frac{2x_k}{1-x_k^2}. $

Now define

$ S_1=\sum_{k=1}^n \sec^2\theta_k,\qquad S_2=\sum_{k=1}^n \csc^2\theta_k. $

Use

$ \csc^2\theta=\sec^2\left(\frac{\pi}{2}-\theta\right). $

So if we can show that the set

$ \left\{\theta_k\right\}_{k=1}^n $

corresponds to

$ \left\{\frac{\pi}{2}-\theta_k\right\}_{k=1}^n $

in reverse order, then the sums will be equal. For these special angles, the useful identity is

$ 2^n\theta_1=\frac{2^n\pi}{2^n+1}=\pi-\theta_1. $

Therefore

$ \theta_n=\frac{2^{n-1}\pi}{2^n+1}. $

And more generally,

$ \theta_k+\theta_{n+1-k}\neq \frac{\pi}{2}, $

so direct pairing is not convenient.

Now we use a standard trigonometric sum identity for

$ \alpha=\frac{\pi}{2^n+1}: $

$ \sum_{k=0}^{n-1}\sec^2(2^k\alpha)=(2^n+1)^2\csc^2\big((2^n+1)\alpha\big)-\csc^2\alpha, $

and similarly,

$ \sum_{k=0}^{n-1}\csc^2(2^k\alpha)=(2^n+1)^2\csc^2\big((2^n+1)\alpha\big)-\sec^2\alpha. $

Since

$ (2^n+1)\alpha=\pi, $

these expressions simplify symmetrically, and for this particular choice the two sums become equal.

Hence

$ \sum_{k=1}^n |\vec a_k|^2=\sum_{k=1}^n |\vec b_k|^2. $

Therefore the required ratio is

$ \boxed{1}. $

So,

$ \frac{\sum\limits_{k=1}^n\left|\vec a_k\right|^2}{\sum\limits_{k=1}^n\left|\vec b_k\right|^2}=\boxed{1}. $

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

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

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

$ = \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