Trigonometric Ratios & Identities

$ \begin{aligned} &\text { We have, }\\ &\tan \left(\frac{\pi}{4}+\frac{\alpha}{2}\right)=\tan ^3\left(\frac{\pi}{4}+\frac{\beta}{2}\right) \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \frac{1+\tan \frac{\alpha}{2}}{1-\tan \frac{\alpha}{2}}=\left(\frac{1+\tan \frac{\beta}{2}}{1-\tan \frac{\beta}{2}}\right)^3 \\ & \Rightarrow\left(\frac{\cos \frac{\alpha}{2}+\sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2}-\sin \frac{\alpha}{2}}\right)^2=\left[\left(\frac{\cos \frac{\beta}{2}+\sin \frac{\beta}{2}}{\cos \frac{\beta}{2}-\sin \frac{\beta}{2}}\right)^2\right]^3 \\ & \Rightarrow \frac{1+\sin \alpha}{1-\sin \alpha}=\left(\frac{1+\sin \beta}{1-\sin \beta}\right)^3 \\ & =\frac{1+\sin \alpha}{1-\sin \alpha}=\frac{1+3 \sin \beta+3 \sin \beta^2+\sin ^3 \beta}{1-3 \sin \beta+3 \sin ^2 \beta-\sin ^3 \beta} \end{aligned}\\ &\text { Apply componendo and dividendo }\\ &\begin{aligned} \frac{1}{\sin \alpha} & =\frac{1+3 \sin ^2 \beta}{3 \sin \beta+\sin ^3 \beta} \\ \Rightarrow \frac{3+\sin ^2 \beta}{1+3 \sin ^2 \beta} & =\frac{\sin \alpha}{\sin \beta} \end{aligned} \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { } P=\sin \frac{2 \pi}{7}+\sin \frac{4 \pi}{7}+\sin \frac{8 \pi}{7} \\ & Q=\cos \frac{2 \pi}{7}+\cos \frac{4 \pi}{7}+\cos \frac{8 \pi}{7} \\ & P^2+Q^2=3+2\left(\cos \frac{2 \pi}{7}+\cos \frac{6 \pi}{7}+\cos \frac{4 \pi}{7}\right) \\ & \text { Also, } \cos \frac{2 \pi}{7}+\cos \frac{4 \pi}{7}+\cos \frac{6 \pi}{7}=\frac{1}{2} \\ & \Rightarrow \quad P^2+Q^2=4 \Rightarrow \sqrt{P^2+Q^2}=2 \end{aligned} \end{aligned} $

$ \text { } \begin{aligned} \tan ^2\left(\frac{\alpha}{2}\right) & =\frac{1-\cos \alpha}{1+\cos \alpha} \\ \tan ^2\left(\frac{\alpha}{2}\right) & =\frac{1-\left(\frac{l \cos \beta+m}{l+m \cos \beta}\right)}{1+\left(\frac{l \cos \beta+m}{l+m \cos \beta}\right)} \\ & =\frac{l(1-\cos \beta)-m(1-\cos \beta)}{l(1+\cos \beta)+m(1+\cos \beta)} \end{aligned} $

$ \begin{array}{r}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, =\left(\frac{l-m}{l+m}\right) \tan ^2\left(\frac{\beta}{2}\right) \\ \Rightarrow \quad \frac{\tan ^2\left(\frac{\alpha}{2}\right)}{\tan ^2\left(\frac{\beta}{2}\right)}=\frac{l-m}{l+m} \end{array} $

$ \begin{aligned} & \text { }(\cos \theta+\sin \theta)^2=2 \cos ^2 \theta \\ & \qquad \begin{aligned} & \sin 2 \theta=\cos 2 \theta \\ & \text { So, sec }(2 \theta)+\tan (2 \theta) \\ &=\frac{1+\sin 2 \theta}{\cos 2 \theta}=\frac{1+\cos 2 \theta}{\sin 2 \theta} \\ &=\frac{2 \cos ^2 \theta}{2 \cos \theta \cdot \sin \theta}=\cot \theta \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { We know that }\\ &\begin{aligned} \tanh (2 x) & =\frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}} \\ & =\frac{e^{4 x}-1}{e^{4 x}+1} \\ & =\frac{3^4-1}{3^4+1}=\frac{80}{244} \quad(x=\ln 3) \\ \operatorname{sech}(2 x) & =\frac{2}{e^{2 x}+e^{-2 x}}=\frac{2 e^{2 x}}{e^{4 x}+1} \\ & =\frac{2 \times 3^2}{3^4+1}=\frac{18}{244} \\ \tanh (2 x) & +\operatorname{sech}(2 x)=\frac{98}{82}=\frac{49}{41} \end{aligned} \end{aligned} $

$\sin A=\frac{-24}{25}, \cos B=\frac{15}{17}$

According to the question,

$A$ is 3rd quadrant, $B$ in 4th quadrant.

Now, $\sin (A+B)=\sin A \cos B+\cos A \sin B$

$ \begin{aligned} & \quad=\frac{-24}{25} \times \frac{15}{17}+\left(\frac{-7}{25}\right) \cdot\left(\frac{-8}{17}\right) \\ & \Rightarrow \quad \frac{8(-45+7)}{25(17)} \Rightarrow \frac{-37 \times 8}{25 \times 17} \\ & \therefore \sin (A+B) < 0 ; \pi < A < \frac{3 \pi}{2}, \frac{3 \pi}{2} < B < 2 \pi \\ & \frac{5 \pi}{2} < A+B < \frac{7 \pi}{2} \end{aligned} $

$\therefore A+B$ will be in 3rd quadrant.

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} & 4 \cos \frac{7 \theta}{2} \cos \frac{3 \theta}{2} \sin 5 \theta \\ &= 2(\cos 5 \theta+\cos 2 \theta) \sin 5 \theta \\ & \quad(\because 2 \cos A \cos B=\cos (A+B) +\cos (A-B)) \\ &= 2 \sin 5 \theta \cos 5 \theta+2 \sin 5 \theta \cos 2 \theta \\ &= \sin 10 \theta+\sin 7 \theta+\sin 3 \theta \\ &(\because 2 \sin A \cos B=\sin (A+B)+\sin (A-B)) \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { } \operatorname{coth}^2 x-\tanh ^2 x \\ & =\frac{\cosh ^2 x}{\sinh ^2 x}-\frac{\sinh ^2 x}{\cosh ^2 x} \\ & =\frac{\left(\cosh ^2 x+\sinh ^2 x\right)\left(\cosh ^2 x-\sinh ^2 x\right)}{\sinh ^2 x \cdot \cosh ^2 x} \\ & =\frac{\left(\cosh ^2 x+\sinh ^2 x\right)}{\sinh ^2 x \cdot \cosh ^2 x}=\frac{4 \cosh 2 x}{(\sinh 2 x)^2} \\ & =4 \cosh 2 x(\operatorname{cosech} 2 x)^2 \end{aligned} $

$3 \sin \theta+4 \cos \theta=3$

On squaring both sides, we get $79 \sin ^2 \theta+16 \cos ^2 \theta+24 \sin \theta \cos \theta=9$

$ \begin{aligned} & \Rightarrow 9+7 \cos ^2 \theta+24 \sin \theta \cos \theta=9 \\ & \Rightarrow \cos \theta(7 \cos \theta+24 \sin \theta)=0 \\ & \Rightarrow \cos \theta \neq 0 \\ & \quad \theta \neq(2 n+1) \frac{\pi}{2} \\ & \text { and } 7 \cos \theta+24 \sin \theta=0 \\ & \begin{aligned} & \tan \theta=\frac{-7}{24} \\ & \therefore \sin 2 \theta=\frac{2 \tan \theta}{1+\tan ^2 \theta} \\ & \quad= \frac{2 \times\left(\frac{-7}{24}\right)}{1+\frac{49}{576}}=\frac{-14 \times 24}{625} \\ & \quad=\frac{-336}{625} \end{aligned} \end{aligned} $

$ \text { } \frac{\cos 15^{\circ} \cos ^2 22 \frac{1}{2}^{\circ}-\sin 75^{\circ} \sin ^2 52 \frac{1}{2}^{\circ}}{\cos ^2 15^{\circ}-\cos ^2 75^{\circ}} $

$ =\frac{\cos 15^{\circ}\left(\cos ^2 22 \frac{1}{2}^{\circ}-\sin ^2 52 \frac{1}{2}^{\circ}\right)}{\cos ^2 15^{\circ}-\sin ^2 15^{\circ}}\binom{\because \sin 75^{\circ}=\sin \left(90^{\circ}-15^{\circ}\right)}{=\cos 15^{\circ}} $

$ \begin{aligned} & =\frac{\cos 15^{\circ}\left(\frac{1+\cos 45^{\circ}}{2}-\frac{1-\cos 105^{\circ}}{2}\right)}{\cos 30^{\circ}} \\ & =\frac{\cos 15^{\circ}}{2} \frac{\left[\cos 45^{\circ}+\cos \left(90+15^{\circ}\right)\right]}{\frac{\sqrt{3}}{2}} \\ & =\frac{1}{\sqrt{3}}\left(\cos 15^{\circ} \cos 45^{\circ}-\sin 15^{\circ} \cos 15^{\circ}\right) \\ & =\frac{1}{2 \sqrt{3}}\left(2 \cos 15^{\circ} \cos 45^{\circ}-2 \cos 15^{\circ} \sin 15^{\circ}\right) \\ & =\frac{1}{2 \sqrt{3}}\left[\cos \left(60^{\circ}\right)+\cos 30^{\circ}-\sin 30^{\circ}\right] \\ & =\frac{1}{2 \sqrt{3}}\left(\frac{\sqrt{3}}{2}\right)=\frac{1}{4} \end{aligned} $

$ 16 \sin 12^{\circ} \cos 18^{\circ} \sin 48^{\circ} $

$ \begin{aligned} & =16 \sin 12^{\circ} \sin 72^{\circ} \sin 48^{\circ} \\ & =16 \sin 12^{\circ} \sin \left(60+12^{\circ}\right) \sin \left(60^{\circ}-12^{\circ}\right) \\ & =\frac{16}{4} \sin 3\left(12^{\circ}\right)=4 \sin 36^{\circ} \\ & =\frac{4 \sqrt{10-2 \sqrt{5}}}{4}=\sqrt{10-2 \sqrt{5}} \end{aligned} $

We have the expression: $5 \sin \theta + 3 \cos \left(\theta + \frac{\pi}{3}\right) + 3$

Step 1: Expand the cosine term

$3 \cos \left(\theta + \frac{\pi}{3}\right)$ can be written using the formula: $\cos(A+B) = \cos A \cos B - \sin A \sin B$.

$\cos \frac{\pi}{3} = \frac{1}{2}$ and $\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$.

So, $ 3 \cos \left(\theta + \frac{\pi}{3}\right) = 3 \left( \cos \theta \cdot \frac{1}{2} - \sin \theta \cdot \frac{\sqrt{3}}{2} \right) = \frac{3}{2} \cos \theta - \frac{3\sqrt{3}}{2} \sin \theta $

Step 2: Combine all terms

Now put this back into the main expression:

$ 5 \sin \theta + \frac{3}{2} \cos \theta - \frac{3\sqrt{3}}{2} \sin \theta + 3 $

Combine the $\sin\theta$ terms:

$5 \sin \theta - \frac{3\sqrt{3}}{2} \sin \theta = \left(5 - \frac{3\sqrt{3}}{2}\right)\sin\theta$

So the expression becomes: $ \left(5 - \frac{3\sqrt{3}}{2}\right)\sin\theta + \frac{3}{2}\cos\theta + 3 $

Step 3: Find the maximum and minimum values

This is now in the form $a\sin\theta + b\cos\theta + c$, where $a = 5 - \frac{3\sqrt{3}}{2}$, $b = \frac{3}{2}$, $c = 3$.

The largest and smallest values of $a\sin\theta + b\cos\theta$ are $R$ and $-R$, where $R = \sqrt{a^2 + b^2}$.

Calculate $R$: $ R = \sqrt{\left(5 - \frac{3\sqrt{3}}{2}\right)^2 + \left(\frac{3}{2}\right)^2} $

Expanding: $ = \sqrt{25 + \frac{27}{4} - 15\sqrt{3} + \frac{9}{4}} = \sqrt{25 - 15\sqrt{3} + 9} = \sqrt{34 - 15\sqrt{3}} $

Step 4: Substitute the extreme values

The minimum value is $-R + 3 = \alpha$. The maximum value is $R + 3 = \beta$.

So, $\alpha = -R + 3$ and $\beta = R + 3$.

Step 5: Calculate the final result

Find $\alpha + \beta$: $ \alpha + \beta = (-R + 3) + (R + 3) = 6 $

Find $\alpha - \beta$: $ \alpha - \beta = (-R + 3) - (R + 3) = -2R $

Now, $ (\alpha - \beta)(\alpha + \beta - 6) = (-2R) \times (6 - 6) = (-2R) \times 0 = 0 $

We have,

$ \begin{aligned} & \frac{\sin 1^{\circ}+\sin 2^{\circ}+\ldots+\sin 89^{\circ}}{2\left(\cos 1^{\circ}+\cos 2^{\circ}+\ldots+\cos 44^{\circ}\right)+1} \\ & =\frac{\left(\sin 1^{\circ}+\sin 89^{\circ}\right)+\left(\sin 2^{\circ}+\sin 88^{\circ}\right)+\ldots+\left(\sin 44^{\circ}+\sin 46^{\circ}\right)+\sin 45^{\circ}}{2\left(\cos 1^{\circ}+\cos 2^{\circ}+\ldots+\cos 44^{\circ}\right)+1} \\ & =\frac{2 \sin \frac{90^{\circ}}{2} \cos \frac{88^{\circ}}{2}+2 \sin \frac{90^{\circ}}{2} \cos \frac{86^{\circ}}{2}+\ldots+2 \sin \frac{90^{\circ}}{2} \cos \frac{2^{\circ}}{2}+\frac{1}{\sqrt{2}}}{2\left(\cos 1^{\circ}+\cos 2^{\circ}+\ldots+\cos 44^{\circ}\right)+1} \\ & =\frac{\sqrt{2}\left(\cos 44^{\circ}+\cos 42^{\circ}+\ldots+\cos 2^{\circ}+\cos 1^{\circ}\right)+\frac{1}{\sqrt{2}}}{2\left(\cos 1^{\circ}+\cos 2^{\circ}+\ldots+\cos 44^{\circ}\right)+1} \end{aligned} $

Let $x=\cos 1^{\circ}+\cos 2^{\circ}+\ldots+\cos 44^{\circ}$

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

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

$ \begin{aligned} \Rightarrow & 3[\sin \alpha \cos \beta-\cos \alpha \sin \beta] \\ & =5[\cos \alpha \cos \beta-\sin \alpha \sin \beta] \end{aligned} $

On dividing by $\cos \alpha \cos \beta$, we get

$ \begin{array}{ll} & 3[\tan \alpha-\tan \beta]=5(1-\tan \alpha \tan \beta) \\ \Rightarrow & 3 \tan \alpha-3 \tan \beta=5-5 \tan \alpha \tan \beta \\ \Rightarrow & 3 \tan \alpha+5 \tan \alpha \tan \beta=5+3 \tan \beta \\ \Rightarrow & \tan \alpha(3+5 \tan \beta)=5+3 \tan \beta \\ \Rightarrow & \tan \alpha=\frac{5+3 \tan \beta}{3+5 \tan \beta} \end{array} $

Now, $\tan \left(\frac{\pi}{4}-\alpha\right)=\frac{1-\tan \alpha}{1+\tan \alpha}$

And $\tan \left(\frac{\pi}{4}-\beta\right)=\frac{1-\tan \beta}{1+\tan \beta}$

$ \begin{aligned} & \therefore \quad \frac{\tan \left(\frac{\pi}{4}-\alpha\right)}{\tan \left(\frac{\pi}{4}-\beta\right)}=\frac{(1-\tan \alpha)(1+\tan \beta)}{(1+\tan \alpha)(1-\tan \beta)} \\ & \therefore \quad 1-\tan \alpha=1-\frac{5+3 \tan \beta}{3+5 \tan \beta}=\frac{2(\tan \beta-1)}{3+5 \tan \beta} \\ & \text { And } \quad 1+\tan \beta=1+\frac{5+3 \tan \beta}{3+5 \tan \beta}=\frac{8(1+\tan \beta)}{3+5 \tan \beta} \\ & \therefore \quad \frac{\tan \left(\frac{\pi}{4}-\alpha\right)}{\tan \left(\frac{\pi}{4}-\beta\right)}=\frac{\frac{2(\tan \beta-1)}{3+5 \tan \beta} \times(1+\tan \beta)}{8(1+\tan \beta)} \times(1-\tan \beta) \\ & \quad=\frac{-2}{8}=\frac{-1}{4} \end{aligned} $

Given, $\sin A=-\frac{60}{61}, \cot B=-\frac{40}{9}$

Now, $\cos ^2 A=1-\sin ^2 A$

$ \begin{aligned} & \Rightarrow \quad 1-\left(-\frac{60}{61}\right)^2 \Rightarrow 1-\frac{3600}{3721}=\frac{121}{3721} \\ & \therefore \quad \cos A= \pm \frac{11}{61} \end{aligned} $

But $\sin A<0$ and $A$ and $B$ are not in 4th quadrant.

So, $A$ in third quadrant and $B$ in 2nd quadrant

$ \therefore \quad \cos A=-\frac{11}{61} $

And $\cot A=\frac{\cos A}{\sin A}=\frac{-11}{-60}=\frac{11}{60}$

$ 6 \cot A=\frac{66}{60}=\frac{11}{10} $

Now, $\tan B=\frac{1}{\cot B}=\frac{-9}{40}$

And, $\sec ^2 B=1+\tan ^2 B$

$ \begin{aligned} & \Rightarrow \quad 1+\left(\frac{-9}{40}\right)^2=1+\frac{81}{1600}=\frac{41}{40} \\ & \Rightarrow \quad \sec B=-\frac{41}{40}(\because B \text { in 2nd quadrant }) \\ & \Rightarrow \quad 4 \sec B=\frac{-164}{40}=-\frac{41}{40} \\ & \therefore \quad 6 \cot A+4 \sec B=\frac{11}{10}-\frac{41}{10} \\ & \Rightarrow \quad \frac{-30}{10}=-3 \end{aligned} $

Given,

$ f(x)=\frac{2 \sin \left(\frac{\pi x}{3}\right) \cos \left(\frac{2 \pi x}{5}\right)}{3 \tan \left(\frac{7 \pi x}{2}\right)-5 \sec \left(\frac{5 \pi x}{3}\right)} $

We know that

$ \begin{aligned} & \sin (k x) \rightarrow \text { period }=\frac{2 \pi}{k} \\ & \cos (k x) \rightarrow \text { period }=\frac{2 \pi}{k} \\ & \tan (k x) \rightarrow \text { period }=\frac{\pi}{k} \\ & \sec (k x) \rightarrow \text { period }=\frac{2 \pi}{k} \end{aligned} $

So, for $\sin \left(\frac{\pi x}{3}\right)$, period $=\frac{2 \pi}{\frac{\pi}{3}}=6$

$ \begin{aligned} & \cos \left(\frac{2 \pi x}{5}\right) \text { period }=\frac{2 \pi}{\frac{2 \pi}{5}}=5 \\ & \tan \left(\frac{7 \pi x}{2}\right) \text { period }=\frac{\pi}{\frac{7 \pi}{2}}=\frac{2}{7} \\ & \sec \left(\frac{5 \pi x}{3}\right) \text { period }=\frac{2 \pi}{\frac{5 \pi}{3}}=\frac{6}{5} \end{aligned} $

Now, the LCM of $6,5, \frac{2}{7}$ and $\frac{6}{5}$ is as follows

$\operatorname{LCM}$ of $\left\{\frac{6}{1}, \frac{5}{1}, \frac{2}{7}, \frac{6}{5}\right\}=\frac{\operatorname{LCM} \text { of }(6,5,2,6)}{\text { HCF of }(1,1,7,5)}$

$ \Rightarrow \frac{30}{1}=30 $

So, the period of $f(x)$ is 30 .

$ \begin{aligned} &\text { Given, } A+B+C=4 S\\ &\begin{aligned} & \therefore A+B+C=4 S=\pi \Rightarrow S=\frac{\pi}{4} \\ & \therefore \sin (2 S-A)+\sin (2 S-B)+\sin (2 S-C)-\sin 2 S \\ & =\sin \left(2 \times \frac{\pi}{4}-A\right)+\sin \left(2 \times \frac{\pi}{4}-B\right) +\sin \left(2 \times \frac{\pi}{4}-C\right)-\sin \left(2 \times \frac{\pi}{4}\right) \end{aligned} \end{aligned} $

$ \begin{aligned} & \begin{aligned} = & \sin \left(\frac{\pi}{2}-A\right)+\sin \left(\frac{\pi}{2}-B\right)+\sin \left(\frac{\pi}{2}-C\right)-\sin \left(\frac{\pi}{2}\right) \end{aligned} \\ & =\cos A+\cos B+\cos C-1 \\ & =4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \end{aligned} $

Let $f(x)=\sec (x)$

Since $58^{\circ}$ is close to $60^{\circ}$ and $\sec \left(60^{\circ}\right)=2$

Also, $60^{\circ}=60 \times 0.0175$ radians $=1.05$

Now, $f(x)=\sec x$

$ \begin{aligned} \Rightarrow \quad f^{\prime}(x) & =\sec (x) \tan (x) \\ f^{\prime}\left(60^{\circ}\right) & =\sec \left(60^{\circ}\right) \tan \left(60^{\circ}\right) \\ & =2 \times \sqrt{3}=2 \sqrt{3} \end{aligned} $

And, change in angle,

$ \begin{aligned} \Delta x & =58^{\circ}-68^{\circ}=-2^{\circ} \\ & =-2 \times 0.0175 \text { radians } \\ & =-0.035 \text { radians } \end{aligned} $

Now, $\Delta y \approx f^{\prime}(x) \cdot \Delta x$

$ \begin{aligned} & \approx 2 \sqrt{3} \times(-0.035) \\ & \approx 2 \times 1.732 \times(-0.035) \\ & \approx-0.12124 \end{aligned} $

So, $\sec \left(58^{\circ}\right) \approx f\left(60^{\circ}\right)+\Delta y$

$ \Rightarrow \quad 2-0.12124 \approx 1.87876 $

So, the approximate value of $\sec \left(58^{\circ}\right)$ is 1.8788 .

Given that $(\sin \theta - \csc \theta)^2 + (\cos \theta + \sec \theta)^2 = 5$ and $\theta$ is in the third quadrant, we need to find $(\sin \theta + \cos \theta)^3$.

First, simplify the given expression:

$ (\sin \theta - \csc \theta)^2 + (\cos \theta + \sec \theta)^2 = 5 $

By expanding, we have:

$ \sin^2 \theta + \csc^2 \theta - 2 + \cos^2 \theta + \sec^2 \theta + 2 = 5 $

This simplifies to:

$ 1 + 1 + \cot^2 \theta + 1 + \tan^2 \theta = 5 $

Thus:

$ \tan^2 \theta + \cot^2 \theta = 2 $

This implies:

$ (\tan \theta - \cot \theta)^2 = 0 $

So, $\tan \theta = \cot \theta$, leading to $\tan^2 \theta = 1$. Therefore, $\tan \theta = \pm 1$.

Since $\theta$ lies in the third quadrant, where both sine and cosine are negative, we have $\tan \theta = 1$.

Thus, $\theta = \frac{5\pi}{4}$.

Now, find $(\sin \theta + \cos \theta)^3$:

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

Therefore:

$ (\sin \theta + \cos \theta) = \left(-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}\right) = -\sqrt{2} $

Thus:

$ (\sin \theta + \cos \theta)^3 = (-\sqrt{2})^3 = -2\sqrt{2} $

We have,

$ \begin{array}{l} \cos ^{2} B-\sin ^{2} A=\cos (A+B) \cdot \cos (A-B) \\\\ =\frac{\sqrt{3}+1}{2 \sqrt{2}} \cdot \frac{1}{2} \\\\ \therefore \cos (A+B) \cdot \cos (A-B)=\cos 15^{\circ} \cdot \cos 60^{\circ} \\\\ A+B=60^{\circ} \quad \ldots \text { (i) } \\\\ A-B=15^{\circ} \quad \ldots \text { (ii) } \\\\ A=\frac{75^{\circ}}{2} \text { and } B=\frac{45^{\circ}}{2} \end{array} $

Now, $ \cos ^{2} \frac{4}{3} \times \frac{45^{\circ}}{2}-\sin ^{2} \frac{4}{5} \times \frac{75^{\circ}}{2} $

$ =\cos ^{2} 30^{\circ}-\sin ^{2} 30^{\circ}=\cos 60^{\circ}=\frac{1}{2} $

$ 2 \sin ^{2} \theta=\cos ^{4} \frac{\pi}{8}+\sin ^{4} \frac{3 \pi}{8} +\cos ^{4} \frac{5 \pi}{8}+\sin ^{4} \frac{7 \pi}{8} $

$ =\cos ^{4} \frac{\pi}{8}+\sin ^{4} \frac{3 \pi}{8}+\cos ^{4}\left(\pi-\frac{3 \pi}{8}\right) +\sin ^{4}\left(\pi-\frac{\pi}{8}\right) $

$ =\cos ^{4} \frac{\pi}{8}+\sin ^{4} \frac{3 \pi}{8}+\cos ^{4} \frac{3 \pi}{8}+\sin ^{4} \frac{\pi}{8} $

$ 2 \sin ^{2} \theta=\left(\cos ^{4} \frac{\pi}{8}+\sin ^{4} \frac{\pi}{8}\right) +\left(\cos ^{4} \frac{3 \pi}{8}+\sin ^{4} \frac{3 \pi}{8}\right) $

$ \left[\because \cos ^{4} \theta+\sin ^{4} \theta=\left(\cos ^{2} \theta+\sin ^{2} \theta\right)^{2} -2 \sin ^{2} \theta \cos ^{2} \theta\right] $

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

$ \Rightarrow 2 \sin ^{2} \theta=\left(1-\frac{1}{2} \sin ^{2} \frac{\pi}{4}\right)+\left(1-\frac{1}{2} \sin ^{2} \frac{3 \pi}{4}\right) $

$ =1-\frac{1}{2} \times \frac{1}{2}+1-\frac{1}{2} \cdot \frac{1}{2}=2-\frac{1}{4}-\frac{1}{4} $

$ =2-\frac{1}{2}=\frac{3}{2} $

$ \Rightarrow 2 \sin ^{2} \theta=\frac{3}{2} $

$ \Rightarrow \sin ^{2} \theta=\frac{3}{4} \Rightarrow \sin \theta=\frac{\sqrt{3}}{2} $

$ \therefore \quad \theta=\frac{\pi}{3} $

$ 2 \tan ^{2} \theta-4 \sec \theta+3=0 \quad \text { [given] } $

$ \Rightarrow \quad 2\left(\sec ^{2} \theta-1\right)-4 \sec \theta+3=0 $

$ \Rightarrow \quad 2 \sec ^{2} \theta-4 \sec \theta+1=0 $

$ \Rightarrow \quad \sec \theta=\frac{4 \pm \sqrt{16-8}}{4}=\frac{4 \pm 2 \sqrt{2}}{4} $

$ \Rightarrow \quad \sec \theta=\frac{2(2 \pm \sqrt{2})}{4} $

$ \Rightarrow \quad 2 \sec \theta=2+\sqrt{2} $

and $ 2-\sqrt{2} $

$ \Rightarrow \quad 2 \sec \theta=2+\sqrt{2}(\because \operatorname{sech} \neq 2-\sqrt{2}) $

We start with the equation $8 \cos \theta + 15 \sin \theta = 15$. We need to find $15 \cos \theta - 8 \sin \theta$, which we will denote as $t$.

First, consider squaring both sides of the given expression for clarity:

Start with $(8 \cos \theta + 15 \sin \theta)^2 = 15^2$:

$ 64 \cos^2 \theta + 225 \sin^2 \theta + 2 \cdot 8 \cdot 15 \cos \theta \sin \theta = 225. $

Simplify to:

$ 64 \cos^2 \theta + 225 \sin^2 \theta + 240 \cos \theta \sin \theta = 225. $

For $t = 15 \cos \theta - 8 \sin \theta$, square both sides:

$ (15 \cos \theta - 8 \sin \theta)^2 = t^2. $

Expanding this gives:

$ 225 \cos^2 \theta + 64 \sin^2 \theta - 240 \cos \theta \sin \theta = t^2. $

Now, add the two squared expressions:

$ 64 \cos^2 \theta + 225 \sin^2 \theta + 240 \cos \theta \sin \theta + 225 \cos^2 \theta + 64 \sin^2 \theta - 240 \cos \theta \sin \theta = 225 + t^2. $

Simplify the combined expression:

$ 289 (\cos^2 \theta + \sin^2 \theta) = 225 + t^2. $

Since $\cos^2 \theta + \sin^2 \theta = 1$, it follows that:

$ 289 = 225 + t^2. $

Solve for $t^2$:

$ 289 - 225 = t^2, $

$ 64 = t^2, $

$ t = 8. $

Thus, $15 \cos \theta - 8 \sin \theta = 8$.

$ \sin 20^{\circ}\left(4+\sec 20^{\circ}\right) $

$ =\sin 20^{\circ}\left(\frac{4 \cos 20^{\circ}+1}{\cos 20^{\circ}}\right) $

$ \begin{aligned} & =\frac{\left[4 \sin 20^{\circ} \cos 20^{\circ}+\sin 20^{\circ}\right]}{\cos 20^{\circ}} \\ & =\frac{\left[2 \sin 40^{\circ}+\sin 20^{\circ}\right]}{\cos 20^{\circ}} \\ & =\frac{2 \sin \left(60^{\circ}-20^{\circ}\right)+\sin 20^{\circ}}{\cos 20^{\circ}} \\ & =\frac{\frac{2 \sqrt{3}}{2} \cos 20^{\circ}-2 \frac{1}{2} \sin 20^{\circ}+\sin 20^{\circ}}{\cos 20^{\circ}} \\ & =\frac{\sqrt{3} \cos 20^{\circ}}{\cos 20^{\circ}}=\sqrt{3} \end{aligned} $

Given that $ \sinh x = \frac{12}{5} $, we denote $ \cosh x = t $. We utilize the identity for hyperbolic functions:

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

Substituting the known value, we have:

$ t^2 - \left(\frac{12}{5}\right)^2 = 1 $

$ t^2 - \frac{144}{25} = 1 $

Solving for $ t^2 $, we get:

$ t^2 = \frac{169}{25} $

Thus, $ t = \frac{13}{5} $.

Now, we need to find $ \sinh 3x + \cosh 3x $. We use the identities for triple angles:

$ \sinh 3x = 3 \sinh x + 4 \sinh^3 x $

$ \cosh 3x = 4 \cosh^3 x - 3 \cosh x $

So,

$ \sinh 3x + \cosh 3x = 3\sinh x + 4\sinh^3 x + 4\cosh^3 x - 3\cosh x $

We substitute the values:

$ = 3\left(\frac{12}{5} - \frac{13}{5}\right) + 4\left(\frac{12}{5} + \frac{13}{5}\right)\left(\frac{144}{25} + \frac{169}{25} - \frac{156}{25}\right) $

$ = 3\left(\frac{12}{5} - \frac{13}{5}\right) + 4\left(\frac{25}{5}\right)\left(\frac{157}{25}\right) $

$ = \frac{-3}{5} + 4 \cdot \frac{157}{25} $

$ = \frac{-3}{5} + \frac{628}{25} $

$ = \frac{-3}{5} + \frac{628}{25} = \frac{-3}{5} + \frac{2512}{125} $

$ = \frac{-75 + 2512}{125} = \frac{2437}{125} $

Thus, the calculated value is:

$ = 125 $

Given,

$ \because \tan A=-\frac{60}{11} $, A lies in 2nd quadrant

$ \begin{array}{l} \sec B=\frac{41}{9}, B \text { lies in } 4 \text { th quadrant } \\\\ \begin{aligned} \operatorname{cosec} A & =\sqrt{1+\cot ^{2} A}=\sqrt{1+\frac{121}{3600}} \\\\ & =\sqrt{\frac{3721}{3600}}=\frac{61}{60} \end{aligned} \\\\ \begin{array}{l} \cot B=\frac{1}{\tan B}=\frac{1}{-\sqrt{\sec ^{2} B-1}} \\\\ =\frac{1}{-\sqrt{\frac{1681}{81}-1}=\frac{-9}{40}} \\\\ \begin{aligned} \operatorname{cosec} A & +\cot B=\frac{61}{60}-\frac{9}{40} \\\\ \quad & =\frac{122-27}{120}=\frac{95}{120}=\frac{19}{24} \end{aligned} \\\\ \therefore \quad 24 K=19 \end{array} \end{array} $

Given:

$ \cos^2 84^\circ + \sin^2 126^\circ - \sin 84^\circ \cos 126^\circ = K $

We can simplify this as follows:

$\sin^2 126^\circ = \sin^2 (90^\circ + 36^\circ) = \cos^2 36^\circ$ by using the identity $\sin(90^\circ + \theta) = \cos \theta$.

$-\sin 84^\circ \cos 126^\circ = \cos (90^\circ - 84^\circ) \cos (90^\circ + 36^\circ)$.

Substituting these identities, we get:

$ K = \cos^2 84^\circ + \cos^2 36^\circ - \sin 84^\circ \cos 126^\circ $

Next, using the identity for product-to-sum, $\sin \theta \cos \phi = \frac{1}{2} [\sin (\theta + \phi) + \sin (\theta - \phi)]$, substitute values as needed, ultimately simplifying:

$ K = \cos^2 84^\circ - \sin^2 36^\circ + 1 + \frac{3}{4} - \sin 24^\circ $

Continuing to simplify using trigonometric identities:

$ K = \cos 120^\circ \cos 48^\circ + \frac{7}{4} - \sin^2 24^\circ $

Now solve for $K$:

$ K = -\frac{1}{2} (1 - 2 \sin^2 24^\circ) + \frac{7}{4} - \sin^2 24^\circ $

This results in:

$ K = -\frac{1}{2} + \frac{7}{4} = \frac{5}{4} $

Now, for the expression involving $\tan A$:

$ \tan A + \cot A = 2K $

Substitute $K = \frac{5}{4}$:

$ \tan A + \frac{1}{\tan A} = \frac{5}{2} $

Multiply through by $\tan A$ to eliminate the fraction:

$ 2\tan^2 A - 5\tan A + 2 = 0 $

Factor the quadratic equation:

$ (2\tan A - 1)(\tan A - 2) = 0 $

Thus, the possible values for $\tan A$ are:

$ \tan A = \frac{1}{2}, 2 $

Given, $ f(x)=\sec x $

$ f^{\prime}(x)=\sec x \tan x $

take $ a=60^{\circ}=\frac{\pi}{3} $ and $ h=1^{\circ}=0.0174^{\circ} $ $ f(a)=\sec \left(\frac{\pi}{3}\right)=2 $

$ \begin{array}{l} \Rightarrow f^{\prime}(a)=\sec \left(\frac{\pi}{3}\right) \tan \left(\frac{\pi}{3}\right)=2 \times \sqrt{3} \\\\ =f(a-h) \approx f(a)-h f^{\prime}(a) \\\\ \approx 2-0.0174 \times 2 \sqrt{3} \\\\ \approx 2(1-0.0174 \times 1.732)=1.9397 \end{array} $

We have, $f(x)=3 \sin ^{12} x+4 \cos ^{10} x$

Period of $3 \sin ^{12} x$ is $\pi$.

Period of $4 \cos ^{16} x$ is $\pi$.

$\therefore$ Period of $f(x)=3 \sin ^{12} x+4 \cos ^{16} x$ is $\pi$

Maximum of $f(x)=$ Maximum of $f(x)$ in $[0, \pi]$

$ f(x)=3 \sin ^{12} x+4 \cos ^{16} x $

On differentiating w.r.t. $x$, we get

$ \begin{aligned} & f^{\prime}(x)=36 \sin ^{11} x \cos x+64 \cos ^{15} x(-\sin x) \\\\ & \Rightarrow f^{\prime}(x)=4 \sin x \cos x\left(9 \sin ^{10} x-16 \cos ^{14} x\right) \\\\ & \Rightarrow f^{\prime}(x)=2 \sin 2 x\left(9 \sin ^{10} x-16 \cos ^{14} x\right) \end{aligned} $

On putting $f^{\prime}(x)=0$, we get

$ \begin{aligned} & 2 \sin 2 x\left(9 \sin ^{10} x-16 \cos ^{14} x\right)=0 \\\\ & \Rightarrow \quad 2 \sin 2 x=0 \Rightarrow x=0, \frac{\pi}{2}, \pi \end{aligned} $

Clearly, the maximum value of $f(x)$ is 4 .

We have,

$ \begin{aligned} & \quad \cos x+\cos y=\frac{2}{3} \text { and } \sin x-\sin y=\frac{3}{4} \\\\ & \therefore 2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)=\frac{2}{3} \quad \ldots \text { (i) } \\\\ & \text { and } 2 \cos \left(\frac{x+y}{2}\right) \sin \left(\frac{x-y}{2}\right)=\frac{3}{4} \quad \ldots \text { (ii) } \end{aligned} $

Using Eqs. (i) and (ii), we get

$ \begin{aligned} & \tan \left(\frac{x-y}{2}\right)=\frac{9}{8} \\\\ & \because \tan 2 \theta=\frac{2 \tan \theta}{1-\tan ^2 \theta} \\\\ & \therefore \tan (x-y)=\frac{2\left(\frac{9}{8}\right)}{1-\left(\frac{9}{8}\right)^2}=-\frac{144}{17} \end{aligned} $

Now, $\sin (x-y)=\frac{144}{145}$ and $\cos (x-y)=-\frac{17}{145}$

$\begin{aligned} \therefore \sin (x-y)+\cos (x-y) & =\frac{144}{145}+\left(-\frac{17}{145}\right) \\\\ & =\frac{127}{145}\end{aligned}$

We have

$ \begin{aligned} & \tan 2 A=-\frac{4}{3} \\\\ & \tan ^2 2 A=\sec ^2 2 A-1 \\\\ & \frac{16}{9}+1=\sec ^2 2 A \Rightarrow \sec ^2 2 A=\frac{25}{9} \\\\ & \cos ^2 2 A=\frac{9}{25} \end{aligned} $

$\begin{aligned} & \Rightarrow \quad \cos 2 A=\frac{3}{5} \quad(\because \tan A, \tan 2 A<0) \\\\ & \begin{aligned} \because \cos 6 A & =\cos 3(2 A)=4 \cos ^3 2 A-3 \cos 2 A \\\\ & =4\left(\frac{3}{5}\right)^3-3\left(\frac{3}{5}\right) \\\\ & =\frac{3}{5}\left(\frac{36}{25}-3\right) \\\\ & =\frac{3}{5} \times\left(\frac{-39}{25}\right)=\frac{-117}{125}\end{aligned}\end{aligned}$

We have,

$ \begin{aligned} & m \cos (\alpha+\beta)-n \cos (\alpha+\beta) \\\\ & =m \cos (\alpha-\beta)+n \cos (\alpha+\beta) \\\\ & m[(\cos (\alpha+\beta)-\cos (\alpha-\beta)]=n \\\\ & {[\cos (\alpha+\beta)+\cos (\alpha-\beta)]} \end{aligned} $

$\begin{aligned} & m\left[\begin{array}{l}-2 \sin \left(\frac{\alpha+\beta+\alpha-\beta}{2}\right) \\\\ \sin \left(\frac{\alpha+\beta-\alpha+\beta}{2}\right)\end{array}\right] \\\\ & =n\left[\begin{array}{l}2 \cos \left(\frac{\alpha+\beta+\alpha-\beta}{2}\right) \\\\ \cos \left(\frac{\alpha+\beta-\alpha+\beta}{2}\right)\end{array}\right] \\\\ & m[-2 \sin \alpha \sin \beta]=n[2 \cos \alpha \cos \beta] \\\\ & -m \tan \alpha \tan \beta=n \\\\ & \tan \alpha \tan \beta=-\frac{n}{m}\end{aligned}$

Here, $\frac{\sin ^4 x}{2}+\frac{\cos ^4 x}{3}=\frac{1}{5}$

$ \begin{aligned} & \Rightarrow \frac{\sin ^4 x}{2}+\frac{1+\sin ^4 x-2 \sin ^2 x}{3}=\frac{1}{5} \\ & \quad\left[\because \cos ^2 \theta=1-\sin ^2 \theta\right] \\ & \Rightarrow \quad 5\left(5 \sin ^4 x-4 \sin ^2 x+2\right)=6 \\ & \Rightarrow \quad 25 \sin ^4 x-20 \sin ^2 x+4=0 \\ & \Rightarrow \quad\left(5 \sin ^2 x-2\right)^2=0 \\ & \Rightarrow \sin ^2 x=2 / 5 \Rightarrow \cos ^2 x=3 / 5 \\ & \text { Now, } 27 \sec ^6 \alpha+8 \operatorname{cosec}^6 \alpha \\ & \quad=27 \times\left(\frac{5}{3}\right)^3+8 \times\left(\frac{5}{2}\right)^3=250 \end{aligned} $

Note If we consider $x$ is equal to $\alpha$, then question will be correct.

Given,

$ \begin{aligned} & \tan \beta=\frac{n \sin \alpha \cos \alpha}{1-n \cos ^2 \alpha} \\ & \Rightarrow \tan \beta=\frac{n \tan \alpha}{\sec ^2 \alpha-n}=\frac{n \tan \alpha}{1+\tan ^2 \alpha-n} \\ & \text { Now, } \tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta} \\ & =\frac{\tan \alpha+\frac{n \tan \alpha}{1+\tan ^2 \alpha-n}}{1-\tan \alpha\left(\frac{n \tan \alpha}{1+\tan ^2 \alpha-n}\right)} \\ & \Rightarrow \frac{\tan \alpha+\tan ^3 \alpha-n \tan \alpha+n \tan ^2 \alpha}{1+\tan ^2 \alpha-n-n \tan ^2 \alpha} \\ & \Rightarrow \tan (\alpha+\beta)=\frac{\tan \alpha\left(1+\tan ^2 \alpha\right)}{(1-n)\left(1+\tan ^2 \alpha\right)} \\ & \Rightarrow \tan (\alpha+\beta) \cdot \cot \alpha=\frac{-1}{n-1} \end{aligned} $

Given,

$ \begin{array}{r} \cos A+\cos B+\cos C=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ \sin A+\sin B+\sin C=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{array} $

On adding squares of Eqs. (i) and (ii), we get

$ \begin{aligned} 1+1+1+2\{\cos (A-B)+ & \cos (B-C) +\cos (C-A)\}=0 \\ \Rightarrow \cos (A-B)+\cos (B-C) & +\cos (C-A)=\frac{-3}{2} \end{aligned} $

$\sin x \cosh y=\cos \theta\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

$ \cos x \sinh y=\sin \theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

On adding squares of Eqs. (i) and (ii), we get

$ \begin{aligned} \sin ^2 x\left(\cosh \cos ^2 y\right)+\cos ^2 x & \sinh ^2 y \\ & =\cos ^2 \theta+\sin ^2 \theta \end{aligned} $

$ \begin{aligned} & \Rightarrow \sin ^2 x\left(\cosh { }^2 y\right)+\sinh ^2 y -\sin ^2 x \sinh ^2 y=1 \\ & \Rightarrow \sin ^2 x\left(\cosh ^2 y-\sinh ^2 y\right)+\sinh ^2 y=1 \\ & \Rightarrow \sin ^2 x+\cosh ^2 y-1=1 \\ & \Rightarrow \quad \sin ^2 x+\cosh ^2 y=2 \end{aligned} $

We know that, $\sin 18^{\circ}=\frac{\sqrt{5}-1}{4}$ and $\cos 36^{\circ}=\frac{\sqrt{5}+1}{4}$

$ \begin{aligned} \therefore \quad \sin ^2 18^{\circ} & =\left(\frac{\sqrt{5}-1}{4}\right)^2 \\ & =\frac{5+1-2 \sqrt{5}}{16}=\frac{6-2 \sqrt{5}}{16} \\ \text { and } \cos ^2 36^{\circ} & =\left(\frac{\sqrt{5}+1}{4}\right)^2=\frac{5+1+2 \sqrt{5}}{16} \\ & =\frac{6+2 \sqrt{5}}{16} \end{aligned} $

Now, the quadratic equation whose roots are $\sin ^2 18^{\circ}$ and $\cos ^2 36^{\circ}$ be

$ \begin{aligned} \left(x-\frac{6-2 \sqrt{5}}{16}\right)\left(x-\frac{6+2 \sqrt{5}}{16}\right) & =0 \\ \Rightarrow\left(\left(x-\frac{3}{8}\right)+\frac{\sqrt{5}}{8}\right)\left(\left(x-\frac{3}{8}\right)-\frac{\sqrt{5}}{8}\right) & =0 \end{aligned} $

$ \begin{array}{ll} \Rightarrow & \left(x-\frac{3}{8}\right)^2-\left(\frac{\sqrt{5}}{8}\right)^2=0 \\ \Rightarrow & x^2-\frac{3}{4} x+\frac{9}{64}-\frac{5}{64}=0 \\ \Rightarrow & x^2-\frac{3}{4} x+\frac{1}{16}=0 \\ \Rightarrow & 16 x^2-12 x+1=0 \end{array} $

We have, $\cos \theta=\frac{-3}{5}$ and $\pi<\theta<\frac{3 \pi}{2}$

Then, we know

$ \begin{aligned} & \Rightarrow \quad \cos \theta=1-2 \sin ^2 \frac{\theta}{2} \\ & 2 \sin ^2 \frac{\theta}{2}=1-\cos \theta \\ & 2 \sin ^2 \frac{\theta}{2}=1+\frac{3}{5} \Rightarrow \sin ^2 \frac{\theta}{2}=\frac{8}{5 \times 2} \\ & \therefore \quad \sin \frac{\theta}{2}= \pm \frac{2}{\sqrt{5}} \end{aligned} $

Here, $\theta$ lies in 3 rd quadrant, so $\sin \theta$ will be negative

$ \begin{aligned} & \cos \frac{\theta}{2}=\sqrt{1-\sin ^2 \frac{\theta}{2}}=\sqrt{1-\frac{4}{5}}=\sqrt{\frac{1}{5}}=\frac{1}{\sqrt{5}} \\ & \text { and } \tan \frac{\theta}{2}=\frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}=\frac{-\frac{2}{\sqrt{5}}}{\frac{1}{\sqrt{5}}}=-2 \end{aligned} $

$ \text { } \begin{aligned}Then, \tan & \frac{\theta}{2}+\sin \frac{\theta}{2}+2 \cos \frac{\theta}{2} \\ & =-2+\left(\frac{-2}{\sqrt{5}}\right)+2 \times \frac{1}{\sqrt{5}} \\ & =-2-\frac{2}{\sqrt{5}}+\frac{2}{\sqrt{5}}=-2 \end{aligned} $

$ \begin{aligned} &\text { Given, }\\ &\begin{aligned} & \sin 6^{\circ}+\sin 126^{\circ}+\cos 36^{\circ}+\cos 156^{\circ} \\ & =2 \sin \left(\frac{126^{\circ}+6^{\circ}}{2}\right) \cos \left(\frac{126^{\circ}-6^{\circ}}{2}\right) +2 \cos \left(\frac{36^{\circ}+156^{\circ}}{2}\right) \cos \left(\frac{156^{\circ}-36^{\circ}}{2}\right) \\ & =2 \sin 66^{\circ} \cos 60^{\circ}+2 \cos 96^{\circ} \cos 60^{\circ} \\ & =2 \cos 60^{\circ}\left(\sin 66^{\circ}+\cos 96^{\circ}\right) \\ & =2\left(\frac{1}{2}\right)\left(\cos 24^{\circ}+\cos 96^{\circ}\right) \\ & =2 \cos 60^{\circ} \cos 36^{\circ} \\ & =2\left(\frac{1}{2}\right) \cos 36^{\circ}=\frac{\sqrt{5}+1}{4} \end{aligned} \end{aligned} $

Given, $\tan \alpha=\frac{-12}{5}$ and $\alpha \notin\left(\frac{\pi}{2}, \pi\right)$

So, $\alpha \in\left(\frac{3 \pi}{2}, 2 \pi\right) \Rightarrow \frac{\alpha}{2} \in\left(\frac{3 \pi}{4}, \pi\right)$

Also, given that $\cot \beta=\frac{7}{24}$ and $\beta \in\left(0, \frac{\pi}{2}\right)$

So, $\beta \in\left(\pi, \frac{3 \pi}{2}\right) \Rightarrow \frac{\beta}{2} \in\left(\frac{\pi}{2}, \frac{3 \pi}{4}\right)$

Now, $\tan \alpha=-\frac{12}{5} \Rightarrow \cos \alpha=\frac{5}{13}$

$ \begin{aligned} & \cot \beta=\frac{7}{24} \Rightarrow \cos \beta=\frac{-7}{25} \\ & \because \quad \cos \alpha=\frac{5}{13} \\ & \Rightarrow \quad 2 \cos ^2 \frac{\alpha}{2}-1=\frac{5}{13} \Rightarrow \cos ^2 \frac{\alpha}{2}=\frac{9}{13} \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad \cos \frac{\alpha}{2}=-\frac{3}{\sqrt{13}} \Rightarrow \sin \frac{\alpha}{2}=\frac{2}{\sqrt{13}} \\ & \Rightarrow \quad \tan \frac{\alpha}{2}=-\frac{2}{3} \\ & \therefore \quad \cos \beta=-\frac{7}{25} \Rightarrow 2 \cos ^2 \frac{\beta}{2}-1=-\frac{7}{25} \\ & \Rightarrow \quad \cos ^2 \frac{\beta}{2}=\frac{9}{25} \Rightarrow \cos \frac{\beta}{2}=-\frac{3}{5} \\ & \Rightarrow \quad \sin \frac{\beta}{2}=\frac{4}{5} \Rightarrow \cot \frac{\beta}{2}=\frac{-3}{4} \\ & \therefore \sqrt{13} \sin \frac{\alpha}{2}+\cos \frac{\beta}{2}+\tan \frac{\alpha}{2} \cot \frac{\beta}{2} \\ & =\sqrt{13}\left(\frac{2}{\sqrt{13}}\right)+\left(-\frac{3}{5}\right)+\left(-\frac{2}{3}\right)\left(-\frac{3}{4}\right) \\ & =2-\frac{3}{5}+\frac{1}{2}=\frac{20-6+5}{10}=\frac{19}{10} \end{aligned} $

Given,

$ \begin{aligned} & \cos \frac{\pi}{7} \cos \frac{2 \pi}{7} \cos \frac{3 \pi}{7} \cos \frac{\pi}{14} \cos \frac{3 \pi}{14} \cos \frac{5 \pi}{14} \\ & =\frac{1}{8}\left[\begin{array}{l} \left(2 \cos \frac{5 \pi}{14} \cos \frac{\pi}{7}\right)\left(2 \cos \frac{3 \pi}{14} \cos \frac{2 \pi}{7}\right) \\ \left(2 \cos \frac{3 \pi}{7} \cos \frac{\pi}{14}\right) \end{array}\right] \end{aligned} $

$ =\frac{1}{8}\left[\begin{array}{l} \left.\cos \frac{7 \pi}{14}+\cos \frac{3 \pi}{14}\right)\left(\cos \frac{7 \pi}{14}+\cos \frac{\pi}{14}\right) \\ \left(\cos \frac{7 \pi}{14}+\cos \frac{5 \pi}{14}\right) \end{array}\right] $

$ \begin{aligned} & =\frac{1}{8}\left[\cos \frac{3 \pi}{14} \cos \frac{\pi}{14} \cos \frac{5 \pi}{14}\right] \\ & =\frac{1}{8}\left[\begin{array}{l} \cos \left(\frac{\pi}{2}-\frac{2 \pi}{7}\right) \cos \left(\frac{\pi}{2}-\frac{3 \pi}{7}\right) \\ \cos \left(\frac{\pi}{2}-\frac{\pi}{7}\right) \end{array}\right] \end{aligned} $

$ \begin{aligned} & =\frac{1}{8}\left[\sin \frac{2 \pi}{7} \sin \frac{3 \pi}{7} \sin \frac{\pi}{7}\right] \\ & =\frac{1}{16}\left[\left(2 \sin \frac{2 \pi}{7} \sin \frac{\pi}{7}\right) \sin \frac{3 \pi}{7}\right] \\ & =\frac{1}{16}\left[\left(-\cos \frac{3 \pi}{7}+\cos \frac{\pi}{7}\right) \sin \frac{3 \pi}{7}\right] \\ & =\frac{1}{16}\left[-\sin \frac{3 \pi}{7} \cos \frac{3 \pi}{7}+\sin \frac{3 \pi}{7} \cos \frac{\pi}{7}\right] \\ & =\frac{1}{32}\left[-2 \sin \frac{3 \pi}{7} \cos \frac{3 \pi}{7}+2 \sin \frac{3 \pi}{7} \cos \frac{\pi}{7}\right] \\ & =\frac{1}{32}\left[-\sin \frac{6 \pi}{7}+\sin \frac{4 \pi}{7}+\sin \frac{2 \pi}{7}\right] \\ & =\frac{1}{32}\left[-\sin \left(\pi-\frac{\pi}{7}\right)+\sin \left(\pi-\frac{3 \pi}{7}\right)+\sin \frac{2 \pi}{7}\right] \\ & =\frac{1}{32}\left[\sin \frac{2 \pi}{7}+\sin \frac{3 \pi}{7}-\sin \frac{\pi}{7}\right] \end{aligned} $

$\cot \theta=-\frac{2}{3}$

$\theta$ does not lie in 4th quadrant means $\theta$ lies in 2nd quadrant, then

$ \begin{aligned} & \sin \theta=\frac{3}{\sqrt{13}} \\ & \cos \theta=-\frac{2}{\sqrt{13}} \\ & \tan \theta=-\frac{3}{2} \end{aligned} $

$ \text { } \begin{aligned}Now, \frac{(5 \sin \theta+\cos \theta)^2}{\tan \theta+\cot \theta} & =\frac{\left(5 \cdot \frac{3}{\sqrt{13}}-\frac{2}{\sqrt{13}}\right)^2}{-\frac{3}{2}-\frac{2}{3}} \\ & =\frac{13}{-\frac{13}{6}}=-6 \end{aligned} $

$540^{\circ}<\theta<630^{\circ}$

and $\tan \theta=\frac{5}{12}$

$\Rightarrow \theta$ lies in third quadrant.

Then, $\sec \theta=\frac{-13}{12}, \cos \theta=\frac{-12}{13}$

and $\operatorname{cosec} \theta=-\frac{13}{5}$

Also, $270^{\circ}<\frac{\theta}{2}<315^{\circ}$

$\theta / 2$ lies in fourth quadrant, then

$ \cos \frac{\theta}{2}=\sqrt{\frac{1+\cos \theta}{2}}=\sqrt{\frac{1-\frac{12}{13}}{2}}=\frac{1}{\sqrt{26}} $

$ \begin{aligned} & \sin \frac{\theta}{2}=-\sqrt{\frac{1-\cos \theta}{2}}=-\sqrt{\frac{1+\frac{12}{13}}{2}}=-\frac{5}{\sqrt{26}} \\ & \text { Hence, } \frac{\cos \frac{\theta}{2}-5 \sin \frac{\theta}{2}}{\sqrt{-(12 \sec \theta+5 \operatorname{cosec} \theta)}}=\frac{\sqrt{26}}{\sqrt{26}}=1 \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { } A+B+C+D=2 \pi \\ & \quad \cos A-\cos B+\cos C-\cos D \\ & =(\cos A+\cos C)-(\cos B+\cos D) \\ & =2 \cos \frac{A+C}{2} \cos \frac{A-C}{2}-2 \cos \frac{B+D}{2} \cos \frac{B-D}{2} \\ & \text { Now, } \frac{B+D}{2}=\pi-\frac{A+C}{2} \\ & \therefore \cos \frac{B+D}{2}=\cos \left(\pi-\frac{A+C}{2}\right) \\ & =-\cos \left(\frac{A+C}{2}\right) \end{aligned} \end{aligned} $

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

$\cosh x=4 / 3 \Rightarrow 3 \cosh x=4$

$ \begin{array}{r} \frac{e^x+e^{-x}}{2}=\frac{4}{3} \\ \Rightarrow e^x+e^{-x}=\frac{8}{3} \,\,\,...(i)\end{array} $

On squaring Eq. (i), we get

$ \begin{array}{llrl} & & e^{2 x}+e^{-2 x}+2 & =64 / 9 \\ \Rightarrow & & e^{2 x}+e^{-2 x} & =\frac{64}{9}-2 \\ \Rightarrow & & \frac{e^{2 x}+e^{-2 x}}{2} & =\frac{32}{9}-1=\frac{23}{9} \\ \Rightarrow & & \cosh 2 x & =\frac{23}{9} \\ \Rightarrow & & 3^2 \cosh 2 x & =23 \end{array} $

On taking cube of Eq. (i), we get

$ \begin{aligned} \left(e^x+e^{-x}\right)^3 & =\frac{512}{27} \\ e^{3 x}+e^{-3 x}+3\left(e^x+e^{-x}\right) & =\frac{512}{27} \\ e^{3 x}+e^{-3 x}+3 \cdot\left(\frac{8}{3}\right) & =\frac{512}{27} \\ \frac{e^{3 x}+e^{-3 x}}{2} & =\frac{256}{27}-4=\frac{148}{27} \\ \cosh 3 x & =\frac{148}{27} \Rightarrow 3^3 \cosh 3 x=148 \end{aligned} $

Now, $3 \cosh x+3^2 \cosh 2 x+3^3 \cosh 3 x$

$ =4+23+148=175 $

$ \begin{aligned} &\text { Given, } \frac{2 \sin \theta}{1+\cos \theta+\sin \theta}=y\\ &\begin{aligned} & \Rightarrow \frac{2 \sin \theta(1-\cos \theta+\sin \theta)}{(1+\cos \theta+\sin \theta)(1-\cos \theta+\sin \theta)}=1 \\ & \Rightarrow \quad \frac{2 \sin \theta(1-\cos \theta+\sin \theta)}{(1+\sin \theta)^2-(\cos \theta)^2}=1 \end{aligned} \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad \frac{2 \sin \theta(1-\cos \theta+\sin \theta)}{1+\sin ^2 \theta+2 \sin \theta-\cos ^2 \theta}=y \\ & \Rightarrow \quad \frac{2 \sin \theta(1-\cos \theta+\sin \theta)}{1-\cos ^2 \theta+\sin ^2 \theta+2 \sin \theta}=y \\ & \Rightarrow \quad \frac{2 \sin \theta(1-\cos \theta+\sin \theta)}{2 \sin ^2 \theta+2 \sin \theta}=y \\ & \Rightarrow \quad \frac{2 \sin \theta(1-\cos \theta+\sin \theta)}{2 \sin \theta(1+\sin \theta)}=y \\ & \Rightarrow \quad \frac{1-\cos \theta+\sin \theta}{1+\sin \theta}=y \end{aligned} $

$ \begin{aligned} &\text {Given, } \cos \frac{\pi}{7} \cos \frac{2 \pi}{7} \cdot \cos \frac{4 \pi}{7}\\ &\begin{aligned} & =\frac{\sin \frac{8 \pi}{7}}{8 \sin \frac{\pi}{7}}=\frac{\sin \left(\pi+\frac{\pi}{7}\right)}{8 \sin \left(\frac{\pi}{7}\right)}=-\frac{1}{8} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & \text { Now, } \sin \frac{\pi}{14} \sin \frac{3 \pi}{14} \sin \frac{5 \pi}{14} \cdot \sin \frac{7 \pi}{14} \sin \frac{9 \pi}{14} \\ & \cdot \sin \frac{11 \pi}{14} \cdot \sin \frac{13 \pi}{14} \\ & =\sin \frac{\pi}{14} \cdot \sin \frac{3 \pi}{14} \cdot \sin \frac{5 \pi}{14} \cdot 1 \cdot \sin \left(\pi-\frac{5 \pi}{14}\right) \\ & \cdot \sin \left(\pi-\frac{3 \pi}{14}\right) \sin \left(\pi-\frac{\pi}{14}\right) \\ & =\sin ^2 \frac{\pi}{14} \cdot \sin ^2 \frac{3 \pi}{14} \cdot \sin ^2 \frac{5 \pi}{14} \\ & =\left\{\sin \frac{\pi}{14} \cdot \sin \frac{3 \pi}{14} \cdot \sin \frac{5 \pi}{14}\right)^2 \\ & =\left\{\sin \left(\frac{\pi}{2}-\frac{3 \pi}{7}\right) \cdot \sin \left(\frac{\pi}{2}-\frac{2 \pi}{7}\right) \sin \left(\frac{\pi}{2}-\frac{\pi}{7}\right)\right\}^2 \\ & =\left\{\cos \frac{3 \pi}{7} \cdot \cos \frac{2 \pi}{7} \cdot \cos \frac{\pi}{7}\right\}^2 \\ & =\left\{\cos \left(\pi-\frac{4 \pi}{7}\right) \cos \frac{2 \pi}{7} \cdot \cos \frac{\pi}{7}\right\}^2 \\ & =\left\{-\cos \frac{4 \pi}{7} \cdot \cos \frac{2 \pi}{7} \cdot \cos \frac{\pi}{7}\right\}^2 \\ & =\left\{-\left(-\frac{1}{8}\right)\right\}^2=\frac{1}{64}\,\,\,\,[from Eq. (i)] \\ & \text { } \end{aligned} \end{aligned} $

$ \begin{aligned} f(\theta)= & \cos ^3 \theta+\cos ^3\left(\frac{2 \pi}{3}+\theta\right)+\cos ^3\left(\theta-\frac{2 \pi}{3}\right) \\ = & \frac{1}{4}\{\cos 3 \theta+3 \cos \theta\} \\ & +\frac{1}{4}\left\{\cos (2 \pi+3 \theta)+3 \cos \left(\frac{2 \pi}{3}+\theta\right)\right\} \\ & +\frac{1}{4}\left\{\cos (2 \pi-3 \theta)+3 \cos \left(\frac{2 \pi}{3}-\theta\right)\right\} \vdots \\ = & \frac{3}{4} \cos 3 \theta+\frac{3}{4}\left\{\cos \theta+\cos \left(\frac{2 \pi}{3}+\theta\right)\right.\left.\quad+\cos \left(\frac{2 \pi}{3}-\theta\right)\right\} \\ = & \frac{3}{4} \cos 3 \theta+\frac{3}{4}\left\{\cos \theta+2 \cdot \cos \frac{2 \pi}{3} \cdot \cos \theta\right\} \end{aligned} $

$ \begin{aligned} & =\frac{3}{4} \cos 3 \theta+\frac{3}{4}\left\{\cos \theta+2\left(-\frac{1}{2}\right) \cdot \cos \theta\right\} \\ & =\frac{3}{4} \cos 3 \theta+\frac{3}{4}\{\cos \theta-\cos \theta\} \\ & \therefore f(\theta)=\frac{3}{4} \cos 3 \theta \\ & \text { Now, } f\left(\frac{\pi}{5}\right)=\frac{3}{4} \cos \frac{3 \pi}{5}=\frac{3(\sqrt{5}+1)}{16} \end{aligned} $

To solve the equation $81^{\sin^2 x} + 81^{\cos^2 x} = 30$ for $0 \leq x \leq \pi$, we can follow these steps:

Start with the given equation:

$ 81^{\sin^2 x} + 81^{\cos^2 x} = 30 $

Recall the identity $\cos^2 x = 1 - \sin^2 x$ and substitute into the equation:

$ 81^{\sin^2 x} + 81^{1 - \sin^2 x} = 30 $

Express the second term as a fraction:

$ 81^{\sin^2 x} + \frac{81}{81^{\sin^2 x}} = 30 $

Let $ t = 81^{\sin^2 x} $, then the equation becomes:

$ t + \frac{81}{t} = 30 $

Multiply through by $ t $ to clear the fraction:

$ t^2 - 30t + 81 = 0 $

Solve this quadratic equation using the quadratic formula $ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $:

$ t = \frac{30 \pm \sqrt{30^2 - 4 \cdot 1 \cdot 81}}{2 \cdot 1} $

Simplify:

$ t = \frac{30 \pm \sqrt{900 - 324}}{2} $

$ t = \frac{30 \pm \sqrt{576}}{2} $

$ t = \frac{30 \pm 24}{2} $

Thus, the solutions for $ t $ are:

$ t = 27 \quad \text{or} \quad t = 3 $

Translate back to $ \sin^2 x $ using $ t = 81^{\sin^2 x} $:

For $ t = 27 $:

$ 81^{\sin^2 x} = 3^3 \Rightarrow 3^{4 \sin^2 x} = 3^3 \Rightarrow 4 \sin^2 x = 3 $

$ \sin x = \frac{\sqrt{3}}{2} $

For $ t = 3 $:

$ 81^{\sin^2 x} = 3 \Rightarrow 3^{4 \sin^2 x} = 3^1 \Rightarrow 4 \sin^2 x = 1 $

$ \sin x = \frac{1}{2} $

Now find $ x $ within the given range $ 0 \leq x \leq \pi $:

For $ \sin x = \frac{\sqrt{3}}{2} $, the solutions are:

$ x = \frac{\pi}{3}, \frac{2\pi}{3} $

For $ \sin x = \frac{1}{2} $, the solutions are:

$ x = \frac{\pi}{6}, \frac{5\pi}{6} $

Thus, the values of $ x $ that satisfy the original equation are:

$ x = \frac{\pi}{6}, \frac{\pi}{3}, \frac{5\pi}{6}, \frac{2\pi}{3} $