Trigonometric Ratio and Identites

Given :

$ \frac{\tan (A-B)}{\tan A}+\frac{\sin ^2 C}{\sin ^2 A}=1 $

$\Rightarrow $ $ \frac{\sin ^2 C}{\sin ^2 A}=1-\frac{\tan (A-B)}{\tan A} $

$\Rightarrow $$ \frac{\sin ^2 C}{\sin ^2 A}=\frac{\tan A-\tan (A-B)}{\tan A}$ .....(1)

We know, $\tan X-\tan Y=\frac{\sin (X-Y)}{\cos X \cos Y}$

$\therefore $ $\tan A-\tan (A-B)=\frac{\sin (A-(A-B))}{\cos A \cos (A-B)}=\frac{\sin B}{\cos A \cos (A-B)}$

Substitute back into (1)

$ \frac{\sin ^2 C}{\sin ^2 A}=\frac{\sin B}{\cos A \cos (A-B) \tan A} $

$\Rightarrow $$\frac{\sin ^2 C}{\sin ^2 A}=\frac{\sin B}{\cos A \cos (A-B) \frac{\sin A}{\cos A}}=\frac{\sin B}{\sin A \cos (A-B)}$

$\Rightarrow $ $\sin ^2 C=\frac{\sin ^2 A \sin B}{\sin A \cos (A-B)}=\frac{\sin A \sin B}{\cos (A-B)}$

Divide by $\cos A \cos B$ :

$ \sin ^2 C=\frac{\tan A \tan B}{\frac{\cos A \cos B+\sin A \sin B}{\cos A \cos B}} $

$\Rightarrow $ $\sin ^2 C=\frac{\tan A \tan B}{1+\tan A \tan B} $

$\Rightarrow $ $ \sin ^2 C=\frac{\tan ^2 C}{1+\tan ^2 C} $

$\Rightarrow $ $ \frac{\tan ^2 C}{1+\tan ^2 C}=\frac{\tan A \tan B}{1+\tan A \tan B}$

$\Rightarrow $ $\tan ^2 C+\tan ^2 C \tan A \tan B=\tan A \tan B+\tan ^2 C \tan A \tan B$

$\Rightarrow $ $\tan ^2 C=\tan A \tan B$

$\therefore $ $\tan A, \tan C, \tan B$ are in G.P.

$ \begin{aligned} &\begin{aligned} & \frac{\sqrt{3} \operatorname{cosec} 20^{\circ}-\sec 20^{\circ}}{\cos 20^{\circ} \cos 40^{\circ} \cos 60^{\circ} \cos 80^{\circ}} \\ & =\frac{\frac{\sqrt{3}}{\sin 20^{\circ}}-\frac{1}{\cos 20^{\circ}}}{\cos 20^{\circ} \cos 40^{\circ} \cos 80^{\circ} \cdot \frac{1}{2}}, \quad\left\{\cos 60^{\circ}=\frac{1}{2}\right\} \\ & =\frac{2\left(\frac{\sqrt{3} \cos 20^{\circ}-\sin 20^{\circ}}{\sin 20^{\circ} \cos 20^{\circ}}\right)}{\cos 20^{\circ} \cos 40^{\circ} \cos 80^{\circ}}, \quad \\ & =\frac{2\left(\sqrt{3} \cos 20^{\circ}-\sin 20^{\circ}\right)}{\frac{\sin 40^{\circ}}{2} \cdot \cos 20^{\circ} \cos 40^{\circ} \cos 80^{\circ}},\left\{\sin 20^{\circ} \cos 20^{\circ}=\frac{\sin 40^{\circ}}{2}\right\} \\ & \text { Numerator } \sqrt{3} \cos 20^{\circ}-\sin 20^{\circ}=2 \cdot\left(\frac{\sqrt{3}}{2} \cos 20^{\circ}-\frac{1}{2} \sin 20^{\circ}\right) \\ & =2\left(\sin 60^{\circ} \cos 20^{\circ}-\cos 60^{\circ} \sin 20^{\circ}\right)=2 \sin (60-20)=2 \sin 40^{\circ} \end{aligned}\\ &\text { Substitute this value }\\ &\begin{aligned} & =\frac{2 \cdot 2 \sin 40^{\circ}}{\frac{\sin 80^{\circ}}{4} \cos 20^{\circ} \cos 80^{\circ}}=\frac{4 \sin 40^{\circ}}{\frac{\sin 160^{\circ}}{8} \cos 20^{\circ}},\left\{\sin 80^{\circ} \cos 80^{\circ}=\frac{\sin 160^{\circ}}{2}\right\} \\ & =\frac{4 \times 2 \sin 20^{\circ} \cdot \cos 20^{\circ}}{\sin (180-160) \cos 20^{\circ}} \times 8=\frac{64 \sin 20^{\circ}}{\sin 20^{\circ}}=64 \end{aligned} \end{aligned} $

$\cot x=\frac{5}{12}, x \in\left(\pi, \frac{3 \pi}{2}\right)$ means 3rd quadrant.

To find the value of:

$ \sin 7 x\left(\cos \frac{13 x}{2}+\sin \frac{13 x}{2}\right)+\cos 7 x\left(\cos \frac{13 x}{2}-\sin \frac{13 x}{2}\right) $

Simplify:

$ =\sin 7 x \cos \frac{13 x}{2}+\sin 7 x \sin \frac{13 x}{2}+\cos 7 x \cos \frac{13 x}{2}-\sin \frac{13 x}{2} \cos 7 x $

Rearranging terms and use $\sin A \cos B-\cos A \sin B=\sin (A-B)$ and $\cos A \cos B+\sin A \sin B=\cos (A-B)$ :

$ \begin{aligned} & =\sin 7 x \cos \frac{13 x}{2}-\sin \frac{13 x}{2} \cos 7 x+\cos 7 x \cos \frac{13 x}{2}+\sin 7 x \sin \frac{13 x}{2} \\ & =\sin \left(7 x-\frac{13 x}{2}\right)+\cos \left(7 x-\frac{13 x}{2}\right) \\ & =\sin \frac{x}{2}+\cos \frac{x}{2} \end{aligned} $

Since $x \in\left(\pi, \frac{3 \pi}{2}\right)$, then $\frac{x}{2} \in\left(\frac{\pi}{2}, \frac{3 \pi}{4}\right)$.

Let $L=\sin \frac{x}{2}+\cos \frac{x}{2}$.

In the second quadrant, $\frac{x}{2}$ makes $L$ positive because $\sin \frac{x}{2}>\left|\cos \frac{x}{2}\right|$.

Squaring:

$ L^2=\sin ^2 \frac{x}{2}+\cos ^2 \frac{x}{2}+2 \sin \frac{x}{2} \cos \frac{x}{2} $

$L^2=1+\sin x=1+\frac{1}{\operatorname{cosec} x}=1+\frac{1}{-\sqrt{1+\cot ^2 x}}$, Since $x$ is in the 3rd quadrant, $\sin x$ must be negative

$ L^2=1+\frac{1}{-\sqrt{1+\frac{25}{144}}}\left(\text { Since } \cot x=\frac{5}{12}\right) $

$ L^2=1-\frac{1}{\sqrt{\frac{169}{144}}}=1-\frac{12}{13}=\frac{1}{13} $

$ L=\frac{1}{\sqrt{13}} $

$\sin \left(\frac{15 \theta}{2}\right)(\cos 8 \theta+\sin 8 \theta)+\cos \left(\frac{15 \theta}{2}\right)(\cos 8 \theta-\sin 8 \theta)$

= $\sin \left(\frac{15 \theta}{2}\right) \cos 8 \theta+\sin \left(\frac{15 \theta}{2}\right) \sin 8 \theta+\cos \left(\frac{15 \theta}{2}\right) \cos 8 \theta-\cos \left(\frac{15 \theta}{2}\right) \sin 8 \theta$

= $\left[\cos \frac{15 \theta}{2} \cos 8 \theta+\sin \frac{15 \theta}{2} \sin 8 \theta\right]+\left[\sin \frac{15 \theta}{2} \cos 8 \theta-\cos \frac{15 \theta}{2} \sin 8 \theta\right]$

= $\cos \left(8 \theta-\frac{15 \theta}{2}\right)+\sin \left(\frac{15 \theta}{2}-8 \theta\right)$

= $\cos \left(\frac{\theta}{2}\right)+\sin \left(-\frac{\theta}{2}\right)=\cos \frac{\theta}{2}-\sin \frac{\theta}{2}$

$\cot \theta=-\frac{1}{2 \sqrt{2}}, \frac{\pi}{2}<\theta<\pi \quad$ (Given)

$\cos \theta=-\frac{1}{3}$

$\begin{aligned} & \cos \theta=1-2 \sin ^2 \frac{\theta}{2} \Longrightarrow-\frac{1}{3}=1-2 \sin ^2 \frac{\theta}{2} \Longrightarrow \sin \frac{\theta}{2}=\sqrt{\frac{2}{3}} \\ & \cos \theta=2 \cos ^2 \frac{\theta}{2}-1 \Longrightarrow-\frac{1}{3}=2 \cos ^2 \frac{\theta}{2}-1 \Longrightarrow \cos \frac{\theta}{2}=\sqrt{\frac{1}{3}}\end{aligned}$

$\therefore $ $\cos \frac{\theta}{2}-\sin \frac{\theta}{2}$

$\begin{aligned} & = \sqrt{\frac{1}{3}}-\sqrt{\frac{2}{3}} \\ & = \frac{1-\sqrt{2}}{\sqrt{3}}\end{aligned}$

$\operatorname{cosec} 10^{\circ}-\sqrt{3} \sec 10^{\circ}$

$\begin{aligned} & =\frac{1}{\sin 10^{\circ}}-\frac{\sqrt{3}}{\cos 10^{\circ}} \\ & =\frac{\cos 10^{\circ}-\sqrt{3} \sin 10^{\circ}}{\sin 10^{\circ} \cos 10^{\circ}} \\ & =4\left[\frac{\frac{1}{2} \cos 10^{\circ}-\frac{\sqrt{3}}{2} \sin 10^{\circ}}{2 \sin 10^{\circ} \cos 10^{\circ}}\right] \\ & =4\left[\frac{\sin \left(30^{\circ}-10^{\circ}\right)}{\sin 20^{\circ}}\right] \\ & =4\end{aligned}$

Let the roots of the equation

$ x^2-2x-5=0 $

be $\tan A$ and $\tan B$.

So, by using relations between roots and coefficients,

$ \tan A+\tan B=2 $

and

$ \tan A \tan B=-5. $

We have to find

$ 20\sin^2\left(\frac{A+B}{2}\right). $

First, find $\tan(A+B)$ using the formula

$ \tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}. $

Substituting the values:

$ \tan(A+B)=\frac{2}{1-(-5)}=\frac{2}{6}=\frac{1}{3}. $

Now let

$ \theta=A+B. $

Then

$ \tan\theta=\frac{1}{3}. $

We need $\sin^2 \frac{\theta}{2}$.

Using the identity

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

we first find $\cos\theta$.

Since

$ \tan\theta=\frac{1}{3}, $

take a right triangle with opposite side $1$, adjacent side $3$. Then hypotenuse is

$ \sqrt{1^2+3^2}=\sqrt{10}. $

So,

$ \cos\theta=\frac{3}{\sqrt{10}}. $

Hence,

$ \sin^2\frac{\theta}{2} = \frac{1-\frac{3}{\sqrt{10}}}{2}. $

Therefore,

$ 20\sin^2\frac{\theta}{2} = 20\cdot \frac{1-\frac{3}{\sqrt{10}}}{2} = 10\left(1-\frac{3}{\sqrt{10}}\right) = 10-\frac{30}{\sqrt{10}}. $

Now simplify:

$ \frac{30}{\sqrt{10}}=3\sqrt{10}. $

So,

$ 20\sin^2\left(\frac{A+B}{2}\right)=10-3\sqrt{10}. $

Therefore, the correct answer is

$ \boxed{10-3\sqrt{10}} $

So, Option C is correct.

$\begin{aligned} & 2\left(\cos ^8 \theta-\sin ^8 \theta\right) \sec 2 \theta=\mathrm{a}^2 \\ & 2\left(\cos ^4 \theta+\sin ^4 \theta\right)\left(\cos ^2 \theta+\sin ^2 \theta\right)\left(\cos ^2 \theta-\sin ^2 \theta\right) \sec 2 \theta=\mathrm{a}^2 \\ & 2\left(\cos ^4 \theta+\sin ^4 \theta\right)=\mathrm{a}^2 \\ & 2\left(1-2 \sin ^2 \theta \cos ^2 \theta\right)=\mathrm{a}^2 \\ & 2\left(1-\frac{\sin ^2 2 \theta}{2}\right)=\mathrm{a}^2, \mathrm{a} \in \mathrm{Z} \\ & \mathrm{a}^2=2-\sin ^2 2 \theta \in[1,2] \\ & \mathrm{a}^2=1 \text { at } \sin ^2 2 \theta=1 \\ & 2 \theta=(2 \mathrm{n}+1) \frac{\pi}{2} \\ & \theta=(2 \mathrm{n}+1) \frac{\pi}{4} \notin \mathrm{P} \\ & \mathrm{n}(\mathrm{S})=0\end{aligned}$

Let

$ K=\sin\left(\frac{\pi}{18}\right)\sin\left(\frac{5\pi}{18}\right)\sin\left(\frac{7\pi}{18}\right) $

We need to find

$ \sin\left(\frac{10K\pi}{3}\right) $

First, write the angles in degrees:

$ \frac{\pi}{18}=10^\circ,\qquad \frac{5\pi}{18}=50^\circ,\qquad \frac{7\pi}{18}=70^\circ $

So,

$ K=\sin 10^\circ \sin 50^\circ \sin 70^\circ $

Now use the standard identity:

$ \sin 3x = 4\sin x \sin(60^\circ+x)\sin(60^\circ-x) $

Taking $x=10^\circ$,

$ \sin 30^\circ = 4\sin 10^\circ \sin 70^\circ \sin 50^\circ $

Since

$ \sin 30^\circ=\frac12 $

we get

$ \frac12 = 4\sin 10^\circ \sin 50^\circ \sin 70^\circ $

Hence,

$ \sin 10^\circ \sin 50^\circ \sin 70^\circ = \frac{1}{8} $

Therefore,

$ K=\frac18 $

Now substitute in the required expression:

$ \sin\left(\frac{10K\pi}{3}\right) = \sin\left(\frac{10}{3}\cdot\frac18\pi\right) = \sin\left(\frac{10\pi}{24}\right) = \sin\left(\frac{5\pi}{12}\right) $

Now,

$ \frac{5\pi}{12}=75^\circ $

So,

$ \sin 75^\circ=\sin(45^\circ+30^\circ) $

Using the formula $\sin(A+B)$:

$ \sin 75^\circ=\sin 45^\circ \cos 30^\circ+\cos 45^\circ \sin 30^\circ $

$ =\frac{1}{\sqrt2}\cdot\frac{\sqrt3}{2}+\frac{1}{\sqrt2}\cdot\frac12 $

$ =\frac{\sqrt3+1}{2\sqrt2} $

So the value is

$ \boxed{\frac{\sqrt3+1}{2\sqrt2}} $

Hence, the correct option is A.

$\begin{aligned} &\text { Let } \phi=\theta+\frac{\pi}{3} \Rightarrow \theta=\phi-\frac{\pi}{6}\\ &\begin{aligned} & x=3 \tan \left(\theta+\frac{\pi}{3}\right)=3 \tan \left(\theta+\frac{\pi}{6}\right) \\ & y=2 \tan \phi \\ & \tan \left(\phi+\frac{\pi}{6}\right)=\frac{\tan \phi+\frac{1}{\sqrt{3}}}{1-\tan \phi \cdot \frac{1}{\sqrt{3}}} \\ & \frac{x}{3}=\frac{\frac{y}{2}+\frac{1}{\sqrt{3}}}{1-\frac{y}{2} \cdot \frac{1}{\sqrt{3}}} \end{aligned} \end{aligned}$

$\begin{aligned} & \Rightarrow \quad x=\frac{3(y \sqrt{3}+2)}{2 \sqrt{3}-y} \\ & x y+\alpha x+\beta y+r=0 \\ & 3\left(\frac{y \sqrt{3}+2}{2 \sqrt{3}-y}\right)+\alpha\left(3 \frac{(y \sqrt{3}+2)}{(2 \sqrt{3}-y)}\right)+\beta y+r=0 \\ & =(3 \sqrt{3}-\beta) y^2+(6+3 \sqrt{3} \alpha+2 \sqrt{3} \beta-y) y \\ & \quad+(6 \alpha+2 \sqrt{3} y)=0 \end{aligned}$

For this identity to hold for all $\theta$, coefficients must be 0

$\begin{aligned} & \therefore \quad \beta=3 \sqrt{3} \\ & \gamma=-\alpha \sqrt{3} \\ & 6+3 \sqrt{3} \alpha+(2 \sqrt{3})(3 \sqrt{3})+\alpha \sqrt{3}=0 \\ & \Rightarrow \alpha=-2 \sqrt{3} \\ & \Rightarrow \beta=6 \\ & \alpha^2+\beta^2+\gamma^2=75 \end{aligned}$

$\begin{aligned} &\begin{aligned} & 10 \sin ^4 \theta+15 \cos ^4 \theta=6 \\ \Rightarrow & 10 \sin ^4 \theta+10 \cos ^4 \theta+5 \cos ^4 \theta=6 \\ \Rightarrow & 10\left[\left(\sin ^2 \theta+\cos ^2 \theta\right)^2-2 \sin ^2 \theta \cos ^2 \theta\right]+5 \cos ^4 \theta=6 \\ \Rightarrow & 10-20\left(1-\cos ^2 \theta\right) \cos ^2 \theta+5 \cos ^4 \theta=6 \end{aligned}\\ &\text { Let } \cos ^2 \theta=x\\ &10-20\left(x-x^2\right)+5 x^2=6 \end{aligned}$

$\begin{aligned} \Rightarrow & 25 x^2-20 x+4=0 \\ & (5 x-2)^2=0 \Rightarrow x=\frac{2}{5} \\ \Rightarrow & \cos ^2 \theta=\frac{2}{5} \Rightarrow \sin ^2 \theta=\frac{3}{5}, \\ & \sec ^2 \theta=\frac{5}{2}, \operatorname{cosec}^2 \theta=\frac{5}{3} \end{aligned}$

$\begin{aligned} \frac{27 \operatorname{cosec}^6 \theta+8 \sec ^6 \theta}{16 \sec ^8 \theta} & =\frac{27\left(\frac{5}{3}\right)^3+8\left(\frac{5}{2}\right)^3}{16\left(\frac{5}{2}\right)^4} \\ & =\frac{5^3+5^3}{5^4}=\frac{2.5^3}{5^4}=\frac{2}{5} \end{aligned}$

$\begin{aligned} &\begin{aligned} & \sin x+\sin ^2 x=1 \\ & \Rightarrow \sin x=\cos ^2 x \Rightarrow \tan x=\cos x \end{aligned}\\ &\therefore \text { Given expression }\\ &\begin{aligned} & =2 \cos ^{12} x+6\left[\cos ^{10} x+\cos ^8 x\right]+2 \cos ^6 x \\ & =2\left[\sin ^6 x+3 \sin ^5 x+3 \sin ^4 x+\sin ^3 x\right] \\ & =2 \sin ^3 x\left[(\sin x+1)^3\right] \\ & =2\left[\sin ^2 x+\sin x\right]^3 \\ & =2 \end{aligned} \end{aligned}$

$\begin{aligned} &\begin{aligned} & \frac{1}{\sin \frac{\pi}{6}} \sum_{r=1}^{13} \frac{\sin \left[\left(\frac{\pi}{4}+\frac{\mathrm{r} \pi}{6}\right)-\left(\frac{\pi}{4}\right)-(\mathrm{r}-1) \frac{\pi}{6}\right]}{\sin \left(\frac{\pi}{4}+(\mathrm{r}-1) \frac{\pi}{6}\right) \sin \left(\frac{\pi}{4}+\frac{\mathrm{r} \pi}{6}\right)} \\ & \frac{1}{\sin \frac{\pi}{6}} \sum_{\mathrm{r}=1}^{13}\left(\cot \left(\frac{\pi}{4}+(\mathrm{r}-1) \frac{\pi}{6}\right)-\cot \left(\frac{\pi}{4}+\frac{\mathrm{r} \pi}{6}\right)\right) \\ & =2 \sqrt{3}-2=\alpha \sqrt{3}+\mathrm{b} \end{aligned}\\ &\text { So } \mathrm{a}^2+\mathrm{b}^2=8 \end{aligned}$

$\begin{aligned} &\begin{aligned} f(x) & =6+16\left(\frac{1}{4} \cos 3 x\right) \sin 3 x \cdot \cos 6 x \\ & =6+4 \cos 3 x \sin 3 x \cos 6 x \\ & =6+\sin 12 x \end{aligned}\\ &\text { Range of } f(x) \text { is [5, 7] }\\ &\begin{aligned} & (\alpha, \beta) \equiv(5,7) \\ & \text { distance }=\left|\frac{15+28+12}{5}\right|=11 \end{aligned} \end{aligned}$

$\begin{aligned} & \sin 70^{\circ}\left(\cot 10^{\circ} \cot 70^{\circ}-1\right) \\ & \Rightarrow \frac{\cos \left(80^{\circ}\right)}{\sin 10}=1 \end{aligned}$

To find the value of $\frac{3 \cos 36^{\circ}+5 \sin 18^{\circ}}{5 \cos 36^{\circ}-3 \sin 18^{\circ}}$ in the form $\frac{a \sqrt{5}-b}{c}$, we need to simplify the given expression. Let's start by using some fundamental trigonometric identities.

We know that:

$\cos 36^{\circ} = \frac{\sqrt{5} + 1}{4}$

$\sin 18^{\circ} = \frac{\sqrt{5} - 1}{4}$

First, substitute these values into the expression:

$\frac{3 \cdot \frac{\sqrt{5} + 1}{4} + 5 \cdot \frac{\sqrt{5} - 1}{4}}{5 \cdot \frac{\sqrt{5} + 1}{4} - 3 \cdot \frac{\sqrt{5} - 1}{4}}$

Simplify the numerator and the denominator:

Numerator: $3 \cdot \frac{\sqrt{5} + 1}{4} + 5 \cdot \frac{\sqrt{5} - 1}{4} = \frac{3(\sqrt{5} + 1) + 5(\sqrt{5} - 1)}{4} = \frac{3\sqrt{5} + 3 + 5\sqrt{5} - 5}{4} = \frac{8\sqrt{5} - 2}{4} = 2 \sqrt{5} - \frac{1}{2}$

Denominator: $5 \cdot \frac{\sqrt{5} + 1}{4} - 3 \cdot \frac{\sqrt{5} - 1}{4} = \frac{5(\sqrt{5} + 1) - 3(\sqrt{5} - 1)}{4} = \frac{5\sqrt{5} + 5 - 3\sqrt{5} + 3}{4} = \frac{2\sqrt{5} + 8}{4} = \frac{\sqrt{5}+4}{2}$

Now combine the simplified numerator and denominator:

$\frac{2 \sqrt{5} - \frac{1}{2}}{\frac{\sqrt{5}+4}{2}} = \frac{(2 \sqrt{5} - \frac{1}{2}) \cdot 2}{\sqrt{5}+4} = \frac{4 \sqrt{5} - 1}{\sqrt{5}+4}$

Rationalizing the denominator by multiplying both the numerator and the denominator by the conjugate of the denominator:

$\frac{(4 \sqrt{5} - 1)(\sqrt{5}-4)}{(\sqrt{5}+4)(\sqrt{5}-4)}$

The denominator simplifies to:

$ \sqrt{5}^2 - 4^2 = 5 - 16 = -11$

The numerator simplifies to:

$ (4 \sqrt{5} - 1)(\sqrt{5}-4) = (4 \sqrt{5} \cdot \sqrt{5} - 4 \cdot 4 \sqrt{5} - 1 \cdot \sqrt{5} + 1 \cdot 4) = (20 - 16 \sqrt{5} - \sqrt{5} + 4) = 24 - 17 \sqrt{5}$

Combining them, we get:

$\frac{24 - 17\sqrt{5}}{-11} = \frac{17\sqrt{5}-24}{11}$

Thus, $a = 17, b = 24, c = 11$.

Therefore, $a + b + c = 17 + 24 + 11 = 52$.

So the correct option is:

Option B: 52

$\begin{aligned} & \sin x=-\frac{3}{5} \text { where } \pi < x < \frac{3 \pi}{2} \\ & \qquad \tan x=\frac{3}{4}, \cos x=\frac{-4}{5} \\ & \therefore 80\left(\tan ^2 x-\cos x\right) \\ & =80\left(\frac{9}{16}+\frac{4}{5}\right) \\ & =80\left(\frac{45+64}{80}\right) \\ & =109 \end{aligned}$

$\begin{aligned} & 4 \cos \theta-3 \sin \theta=1 \\ & 4 \cos \theta-1=3 \sin \theta \\ & 16 \cos ^2 \theta+1-8 \cos \theta=9\left(1-\cos ^2 \theta\right) \\ & \Rightarrow 25 \cos ^2 \theta-8 \cos \theta-8=0 \\ & \Rightarrow \cos \theta=\frac{8 \pm \sqrt{64+4 \times 25 \times 8}}{2.25} \\ & =\frac{8 \pm 4 \sqrt{4+50}}{2.25} \end{aligned}$

$\begin{aligned} & =\frac{4 \pm 2 \sqrt{54}}{25} \\ & \text { As } \theta \in\left[0, \frac{\pi}{4}\right] \\ & \Rightarrow \cos \theta=\frac{4+6 \sqrt{6}}{25}=\frac{4}{3 \sqrt{6}-2} \end{aligned}$

To find the sum of two angles in terms of tangent, we can use the tangent addition formula:

$ \tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A \cdot \tan B} $

Let's compute $\tan (A + B)$ using the given $\tan A$ and $\tan B$:

$ \tan A = \frac{1}{\sqrt{x(x^2+x+1)}} $

$ \tan B = \frac{\sqrt{x}}{\sqrt{x^2+x+1}} $

Now, we can apply the addition formula:

$ \tan (A + B) = \frac{\frac{1}{\sqrt{x(x^2+x+1)}} + \frac{\sqrt{x}}{\sqrt{x^2+x+1}}}{1 - \frac{1}{\sqrt{x(x^2+x+1)}} \cdot \frac{\sqrt{x}}{\sqrt{x^2+x+1}}} $

$ \tan (A + B) = \frac{\frac{1 + \sqrt{x^2}}{\sqrt{x(x^2+x+1)}}}{1 - \frac{1}{(x^2+x+1)}} $

$ \tan (A + B) = \frac{\frac{\sqrt{x^2} + 1}{\sqrt{x(x^2+x+1)}}}{\frac{(x^2+x+1) - 1}{(x^2+x+1)}} $

$ \tan (A + B) = \frac{(x + 1)\sqrt{(x^2+x+1)}}{(x^2 + x){\sqrt{x}}} $

$ \tan (A + B) = \frac{{(x + 1)}\sqrt{(x^2+x+1)}}{x{(x + 1)}{\sqrt{x}}} $

$ \tan (A + B) = \frac{\sqrt{x^2+x+1}}{x^{3/2}} $

$\tan \mathrm{C} = \sqrt{x^{-3}+x^{-2}+x^{-1}} = \sqrt{\frac{1}{x^3}+\frac{1}{x^2}+\frac{1}{x}} = \sqrt{\frac{1+x+x^2}{x^3}} = \frac{\sqrt{x^2+x+1}}{x^{3/2}}.$

We see that:

$ \tan (A + B) = \tan C $

Since tangent is positive and all the angles $A$, $B$, and $C$ are in the first quadrant, we can say that:

$ A + B = C $

Hence, the answer is:

Option A : $C$

Take $e^{\sin x}=t(t>0)$

$\begin{aligned} & \Rightarrow \mathrm{t}-\frac{2}{\mathrm{t}}=2 \\ & \Rightarrow \frac{\mathrm{t}^2-2}{\mathrm{t}}=2 \\ & \Rightarrow \mathrm{t}^2-2 \mathrm{t}-2=0 \\ & \Rightarrow \mathrm{t}^2-2 \mathrm{t}+1=3 \\ & \Rightarrow(\mathrm{t}-1)^2=3 \\ & \Rightarrow \mathrm{t}=1 \pm \sqrt{3} \\ & \Rightarrow \mathrm{t}=1 \pm 1.73 \\ & \Rightarrow \mathrm{t}=2.73 \text { or }-0.73 \text { (rejected as } \mathrm{t}>0) \\ & \Rightarrow \mathrm{e}^{\sin \mathrm{x}}=2.73 \\ & \Rightarrow \log _{\mathrm{e}} \mathrm{e}^{\sin \mathrm{x}}=\log _{\mathrm{e}} 2.73 \\ & \Rightarrow \sin \mathrm{x}=\log _{\mathrm{e}} 2.73>1 \end{aligned}$

So no solution.

To find the value of $k$, the given conditions are:

$3 \sin (\alpha+\beta)=2 \sin (\alpha-\beta)$

And $\tan \alpha = k \tan \beta$

For the first equation, using the sum and difference formulas for sine, we can rewrite the equation as:

$3(\sin \alpha \cos \beta + \cos \alpha \sin \beta) = 2(\sin \alpha \cos \beta - \cos \alpha \sin \beta)$

Simplifying this, we get:

$3 \sin \alpha \cos \beta + 3 \cos \alpha \sin \beta = 2 \sin \alpha \cos \beta - 2 \cos \alpha \sin \beta$

Rearranging the terms, we obtain:

$5 \sin \beta \cos \alpha = - \sin \alpha \cos \beta$

Dividing both sides by $\sin \alpha \cos \beta$, we get:

$\frac{5 \sin \beta \cos \alpha}{\sin \alpha \cos \beta} = -1$

Which simplifies to:

$5 \tan \beta = - \tan \alpha$

So, taking the reciprocal, we have:

$\tan \alpha = -5 \tan \beta$

Therefore, by comparing this equation with the given $\tan \alpha = k \tan \beta$, we find that $k = -5$.

Thus, the value of $k$ is $-5$.

$\tan 15^\circ + \tan 15^\circ - \tan 15^\circ + \tan 15^\circ $

$ = 2\tan 15^\circ $

$ = 2\left( {2 - \sqrt 3 } \right) = 2a \Rightarrow a = 2 - \sqrt 3 $

$\therefore$ ${1 \over a} + a \Rightarrow \left( {2 + \sqrt 3 } \right) + \left( {2 - \sqrt 3 } \right) = 4$

$2\sin {\pi \over {22}}\sin {{3\pi } \over {22}}\sin {{5\pi } \over {22}}\sin {{7\pi } \over {22}}\sin {{9\pi } \over {22}}$

$ = 2\sin \left( {{{11\pi - 10\pi } \over {22}}} \right)\sin \left( {{{11\pi - 8\pi } \over {22}}} \right)\sin \left( {{{11\pi - 6\pi } \over {22}}} \right)\sin \left( {{{11\pi - 4\pi } \over {22}}} \right)\sin \left( {{{11\pi - 2\pi } \over {22}}} \right)$

$ = 2\cos {\pi \over {11}}\cos {{2\pi } \over {11}}\cos {{3\pi } \over {11}}\cos {{4\pi } \over {11}}\cos {{5\pi } \over {11}}$

$ = {{2\sin {{32\pi } \over {11}}} \over {{2^5}\sin {\pi \over {11}}}}$

$ = {1 \over {16}}$

Given that $\alpha = \sin 36^\circ$, we need to determine which equation it is a root of.

We start with the known relationship for $\cos 72^\circ$:

$ \cos 72^\circ = \frac{\sqrt{5}-1}{4} $

Using the double-angle formula for cosine:

$ \cos 72^\circ = 1 - 2 \sin^2 36^\circ $

Substitute $\alpha$ for $\sin 36^\circ$:

$ 1 - 2\alpha^2 = \frac{\sqrt{5}-1}{4} $

Multiply both sides by 4:

$ 4 - 8\alpha^2 = \sqrt{5} - 1 $

Add 1 to both sides:

$ 5 - 8\alpha^2 = \sqrt{5} $

Square both sides to eliminate the radical:

$ (5 - 8\alpha^2)^2 = 5 $

Expand the left side:

$ 25 + 64\alpha^4 - 80\alpha^2 = 5 $

Simplify by subtracting 5 from both sides:

$ 64\alpha^4 - 80\alpha^2 + 20 = 0 $

Divide the entire equation by 4:

$ 16\alpha^4 - 20\alpha^2 + 5 = 0 $

Thus, the equation $16\alpha^4 - 20\alpha^2 + 5 = 0$ is the one for which $\alpha = \sin 36^\circ$ is a root.

$\cos {{2\pi } \over 7} + \cos {{4\pi } \over 7} + \cos {{6\pi } \over 7} = {{\sin 3\left( {{\pi \over 7}} \right)} \over {\sin {\pi \over 7}}}\cos {{\left( {{{2\pi } \over 7} + {{6\pi } \over 7}} \right)} \over 2}$

$ = {{\sin \left( {{{3\pi } \over 7}} \right)\,.\,\cos \left( {{{4\pi } \over 7}} \right)} \over {\sin \left( {{\pi \over 7}} \right)}}$

$ = {{2\sin {{4\pi } \over 7}\cos {{4\pi } \over 7}} \over {2\sin {\pi \over 7}}}$

$ = {{\sin \left( {{{8\pi } \over 7}} \right)} \over {2\sin {\pi \over 7}}} = {{ - \sin {\pi \over 7}} \over {2\sin {\pi \over 7}}} = {{ - 1} \over 2}$

$16\sin 20^\circ \,.\,\sin 40^\circ \,.\,\sin 80^\circ $

$ = 4\sin 60^\circ $ {$\because$ $4\sin \theta \,.\,\sin (60^\circ - \theta )\,.\,\sin (60^\circ + \theta ) = \sin 3\theta $}

$ = 2\sqrt 3 $

$2\sin 12^\circ - \sin 72^\circ $

$ = \sin 12^\circ + ( - 2\cos 42^\circ \,.\,\sin 30^\circ )$

$ = \sin 12^\circ - \cos 42^\circ $

$ = \sin 12^\circ - \sin 48^\circ $

$ = 2\sin 18^\circ \,.\,\cos 30^\circ $

$ = - 2\left( {{{\sqrt 5 - 1} \over 4}} \right)\,.\,{{\sqrt 3 } \over 2}$

$ = {{\sqrt 3 \left( {1 - \sqrt 5 } \right)} \over 4}$