Trigonometric Ratios & Identities

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

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} $

Given, $\sin \alpha+\cos \alpha=m$

$ \begin{aligned} & \Rightarrow \sin ^2 \alpha+\cos ^2 \alpha+2 \sin \alpha \cos \alpha=m^2 \\ & \sin \alpha \cos \alpha=\frac{m^2-1}{2} \\ & \sin ^6 \alpha+\cos ^6 \alpha=\left(\sin ^2 \alpha\right)^3+\left(\cos ^2 \alpha\right)^3 \end{aligned} $

$ \begin{aligned} & =\left(\sin ^2 \alpha+\cos ^2 \alpha\right)\left(\sin ^4 \alpha+\cos ^4 \alpha\right. \\ & -\sin ^2 \alpha \cos ^2 \alpha \text { ) } \\ & =\left(\sin ^2 \alpha\right)^2+\left(\cos ^2 \alpha\right)^2+2 \sin ^2 \alpha \cos ^2 \alpha \\ & -2 \sin ^2 \alpha \cos ^2 \alpha-\sin ^2 \alpha \cos ^2 \alpha \\ & =\left(\sin ^2 \alpha+\cos ^2 \alpha\right)^2-3 \sin ^2 \alpha \cos ^2 \alpha \\ & =1-\frac{3\left(m^2-1\right)^2}{4} \\ & =\frac{4-3\left(m^2-1\right)^2}{4} \end{aligned} $

$\sin A=\frac{-7}{25}, \cos B=\frac{8}{17}$

$A$ does not lie in 3rd quadrant and $B$ does not lie in 1st quadrant.

$\because \sin A=(-\mathrm{ve})$ and $\cos B=(+\mathrm{ve})$

$\therefore A$ lies in 4 th quadrant and $B$ also lies in 4th quadrant.

$ \begin{array}{ll} & \sin A=\frac{-7}{25}=\frac{P_1}{H_1} \Rightarrow \text { Base }=24 \\ \therefore & \tan A=-\frac{P_1}{\text { Base }} \Rightarrow \tan A=\frac{-7}{24} \\ & \cos B=\frac{8}{17}=\frac{\text { Base }}{H_1} \Rightarrow P_1=15 \\ \therefore & \cot B=\frac{- \text { Base }}{P_1} \Rightarrow \cot B=\frac{-8}{15} \end{array} $

Therefore, $8 \tan A-5 \cot B$

$ \begin{aligned} & =8 \times\left(\frac{-7}{24}\right)-5\left(\frac{-8}{15}\right) \\ & =\frac{-56}{24}+\frac{40}{15}=-\frac{7}{3}+\frac{8}{3}=\frac{1}{3} \end{aligned} $

$ \begin{aligned} & \text { } \sin \theta-\cos \theta=\frac{1}{\sqrt{3}} \\ & \sin ^2 \theta+\cos ^2 \theta-2 \sin \theta \cos \theta=\frac{1}{3} \\ & \Rightarrow \quad 1-\sin 2 \theta=\frac{1}{3} \\ & \Rightarrow \quad \sin 2 \theta=\frac{2}{3} \end{aligned} $

Now, $\cos 4 \theta=1-2 \sin ^2 2 \theta$

$ \begin{aligned} & =1-2\left(\frac{2}{3}\right)^2=1-\frac{8}{9}=\frac{1}{9} \\ \sin 6 \theta & =3 \sin 2 \theta-4 \sin ^3 2 \theta \\ & =3\left(\frac{2}{3}\right)-4\left(\frac{2}{3}\right)^3 \\ & =2-4 \cdot \frac{8}{27}=2-\frac{32}{27}=\frac{22}{27} \end{aligned} $

$ \begin{aligned}& \therefore \sin (2 \theta)+\cos (4 \theta)+\sin (6 \theta) & \\ & =\frac{2}{3}+\frac{1}{9}+\frac{22}{27} \\ & =\frac{2 \times 9}{27}+\frac{3}{27}+\frac{22}{27} \\ & =\frac{18+3+22}{27}=\frac{43}{27} \end{aligned} $

$a \tan \alpha+b \tan \beta$

$ \begin{aligned} = & (a+b) \tan \left(\frac{\alpha+\beta}{2}\right) \\ a \tan \alpha+b \tan \beta & \\ = & a \tan \left(\frac{\alpha+\beta}{2}\right)+b \tan \left(\frac{\alpha+\beta}{2}\right) \\ \Rightarrow a \tan \alpha & -a \tan \left(\frac{\alpha+\beta}{2}\right) \\ = & b \tan \left(\frac{\alpha+\beta}{2}\right)-b \tan \beta \\ \Rightarrow a[\tan \alpha & \left.-\tan \left(\frac{\alpha+\beta}{2}\right)\right] \\ = & b\left[\tan \left(\frac{\alpha+\beta}{2}\right)-\tan \beta\right] \end{aligned} $

Using the formula

$ \begin{aligned} & \tan P-\tan Q=\frac{\sin (P-Q)}{\cos P \cdot \cos Q} \\ & \Rightarrow \quad a \cdot \frac{\sin \left(\alpha-\frac{\alpha+\beta}{2}\right)}{\cos \alpha \cos \left(\frac{\alpha+\beta}{2}\right)} \end{aligned} $

$ \begin{aligned} & \text { } \frac{5 \sinh 2 x}{7+6 \cosh 2 x}=\frac{3}{2} \\ & \Rightarrow \quad \frac{5 \cdot \frac{2 \tanh x}{1-\tanh ^2 x}}{7+6\left(\frac{1+\tanh ^2 x}{1-\tanh ^2 x}\right)}=\frac{3}{2} \\ & \Rightarrow \frac{10 \tanh x}{7-7 \tanh ^2 x+6+6 \tanh ^2 x}=\frac{3}{2} \\ & \Rightarrow \quad \frac{10 \tanh x}{13-\tanh ^2 x}=\frac{3}{2} \\ & \Rightarrow 20 \tanh ^2=39-3 \tanh { }^2 x \\ & \Rightarrow 3 \tanh ^2 x+20 \tanh x=39 \end{aligned} $

Given, $\sin (A+B) \sin (A-B)$

$ +\cos (A+B) \cos (A-B)=\frac{1}{2} $

where, $0 < B < \frac{\pi}{2}$

Using $\cos \left(x^2-y\right)=\sin x \sin y+\cos x \cos y$, we get

$ \begin{array}{ll} \cos & \{(A+B)-(A-B)\}=\frac{1}{2} \\ \Rightarrow & \cos 2 B=\frac{1}{2} \\ \Rightarrow & 2 B=\frac{\pi}{3} \Rightarrow B=\frac{\pi}{6} \end{array} $

$ \begin{aligned} &\text { We have, } \frac{1}{\sin 250^{\circ}}+\frac{\sqrt{3}}{\cos 290^{\circ}}\\ &\begin{aligned} & =\frac{1}{\sin \left(180^{\circ}+70^{\circ}\right)}+\frac{\sqrt{3}}{\cos \left(360^{\circ}-70^{\circ}\right)} \\ & =\frac{-1}{\sin 70^{\circ}}+\frac{\sqrt{3}}{\cos 70^{\circ}} \\ & \quad[\because \sin (\pi+\theta)=-\sin \theta \\ & \quad \text { and } \cos (2 \pi-\theta)=\cos \theta] \\ & =\frac{\sqrt{3} \sin 70^{\circ}-\cos 70^{\circ}}{\sin 70^{\circ} \cos 70^{\circ}} \end{aligned} \end{aligned} $

$ \begin{aligned} & =\frac{2 \times 2\left[\sqrt{3} / 2 \sin 70^{\circ}-1 / 2 \cos 70^{\circ}\right]}{2 \sin 70^{\circ} \cos 70^{\circ}} \\ & =\frac{4\left[\cos 30^{\circ} \sin 70^{\circ}-\sin 30^{\circ} \cos 70^{\circ}\right]}{\sin 140^{\circ}} \\ & {[\because 2 \sin x \cos x=\sin 2 x]} \\ & =\frac{4 \sin \left(70^{\circ}-30^{\circ}\right)}{\sin 140^{\circ}} \\ & {[\because \sin (x-y)=\sin x \cos x-\cos x \sin y]} \\ & =\frac{4 \sin 40^{\circ}}{\sin (180-40)}=\frac{4 \sin 40}{\sin 40}=4 \end{aligned} $

$ \begin{aligned} &\text { If } A+B+C=\frac{\pi}{2} \text {, then }\\ &\begin{aligned} & \sqrt{2} \cos \left(\frac{\pi}{4}-A\right)+\sqrt{2} \cos \left(\frac{\pi}{4}-B\right) \\ & +\sqrt{2} \cos \left(\frac{\pi}{4}-C\right)+1 \\ & \begin{array}{r} \sqrt{2}\left[\cos \left(\frac{\pi}{4}-A\right)+\cos \left(\frac{\pi}{4}-B\right)\right] \\ +\sqrt{2}\left(\frac{1}{\sqrt{2}} \cos C+\frac{1}{\sqrt{2}} \sin C\right)+1 \\ {[\because \cos (x-y)=\cos x \cos y+\sin x \sin y]} \end{array} \\ & =2 \sqrt{2} \cos \left(\frac{\frac{\pi}{2}-A-B}{2}\right) \cos \left(\frac{B-A}{2}\right) \\ & +\cos C+\sin C+1 \end{aligned} \end{aligned} $

$ \left[\begin{array}{rl} \because \cos x & +\cos y \\ & =2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right) \end{array}\right] $

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

$ \begin{aligned} & \because A+B+C=\frac{\pi}{2} \text { (given) } \\ & \cos 2 A=2 \cos ^2 A-1 \\ & \text { and } \sin 2 A=2 \sin A \cos A] \\ & =2 \cos \frac{C}{2}\left[\begin{array}{c} \sqrt{2} \cos \left(\frac{B-A}{2}\right) \\ +\cos \frac{C}{2}+\sin \frac{C}{2} \end{array}\right] \\ & =2 \cos \frac{C}{2}\left[\begin{array}{c} \sqrt{2} \cos \left(\frac{B-A}{2}\right)+\sqrt{2} \left.\left(\frac{1}{\sqrt{2}} \cos \frac{C}{2}+\frac{1}{\sqrt{2}} \sin \frac{C}{2}\right)\right] \end{array}\right. \end{aligned} $

$ \begin{gathered} =2 \sqrt{2} \cos \frac{C}{2}\left[\cos \left(\frac{B-A}{2}\right)+\cos \left(\frac{\pi}{4}-\frac{C}{2}\right)\right] \\ {[\because \cos (x-y)=\cos x \cos y+\sin x \sin y]} \\ =2 \sqrt{2} \cos \frac{C}{2}\left[\cos \left(\frac{B-A}{2}\right)+\cos \left(\frac{A+B}{2}\right)\right] \\ {\left[\because A+B+C=\frac{\pi}{2} \text { (given) }\right]} \end{gathered} $

$ \begin{aligned} & =2 \sqrt{2} \cos \frac{C}{2}\left[2 \cos \frac{B}{2} \cos \frac{A}{2}\right] \\ & =4 \sqrt{2} \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2} \end{aligned} $

Given, $\sinh x=\tan A$

$ \begin{aligned} & \Rightarrow \quad\left(\frac{e^x-e^{-x}}{2}\right)=\tan A \\ & \text { Now, }|\tanh x|=\left|\frac{\sinh x}{\cosh x}\right|=\left|\frac{\frac{1}{2}\left(e^x-e^{-x}\right)}{\frac{1}{2}\left(e^x+e^{-x}\right)}\right| \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\left|\frac{\frac{e^x-e^{-x}}{2}}{\sqrt{\left(\frac{e^x-e^{-x}}{2}\right)^2+1}}\right| \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, =\left|\frac{\tan A}{\sqrt{\tan 2} A+1}\right|=\left|\frac{\tan A}{\sqrt{\sec ^2 A}}\right| \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \left.\therefore \quad \mid \because \tan ^2 x+1=\sec ^2 x\right] \\ & \therefore \quad|\tanh x|=\left|\frac{\tan A}{\sec A}\right|=|\sin A| \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} & \frac{\sinh (x+y)+\sinh (x-y)}{\cosh (x+y)-\cosh (x-y)} \\ & =\frac{\frac{1}{2}\left(e^{x+y}-e^{-x-y}\right)+\frac{1}{2}\left(e^{x-y}-e^{-x+y}\right)}{\frac{1}{2}\left(e^{x+y}+e^{-x-y}\right)-\frac{1}{2}\left(e^{x-y}+e^{-x+y}\right)} \\ & =\frac{\left(e^{x+y}+e^{x-y}\right)-\left(e^{-x-y}+e^{-x+y}\right)}{\left(e^{x+y}-e^{x-y}\right)+\left(e^{-x-y}-e^{-x+y}\right)} \\ & =\frac{\left(e^x-e^{-x}\right)\left(e^y+e^{-y}\right)}{\left(e^x-e^{-x}\right)\left(e^y-e^{-y}\right)} \\ & =\frac{\left(\frac{e^y+e^{-y}}{2}\right)}{\left(\frac{e^y-e^{-y}}{2}\right)} \\ & =\frac{\cosh y}{\sinh y}=\operatorname{coth} y \end{aligned} \end{aligned} $

Period of $3 \sin \frac{\pi}{3} x$ is $\frac{2 \pi}{\pi / 3}=6$

Period of $\cos \frac{\pi x}{2}$ is $\frac{2 \pi}{\pi / 2}=4$

Period of $\tan \frac{\pi x}{4}$ is $\frac{\pi}{\pi / 4}=4$

$\alpha=$ Period of

$ \begin{aligned} 3 \sin \frac{\pi}{3} x-\cos \frac{\pi}{2} x+\tan \frac{\pi}{4} x & \\ & =\text { LCM of } 6,4,4=12 \end{aligned} $

Period of $\sin ^2\left(\frac{\pi}{7}+\frac{x}{4}\right)$ is $\frac{\pi}{1 / 4}=4 \pi$

Period of $\sin ^2\left(\frac{\pi}{7}-\frac{x}{4}\right)$ is $\frac{\pi}{1 / 4}=4 \pi$

$ \begin{aligned} & \beta=\text { Period of } \sin ^2\left(\frac{\pi}{7}+\frac{x}{4}\right)-\sin ^2\left(\frac{\pi}{7}-\frac{x}{4}\right) \\ & =\text { LCM of } 4 \pi \text { and } 4 \pi=4 \pi \\ & \begin{array}{r} \cos ^4 x+\sin ^4 x=\left(\cos ^2 x+\sin ^2 x\right)^2 \\ -2 \sin ^2 x \cos ^2 x \end{array} \end{aligned} $

$ \begin{aligned} & \quad=1-\frac{1}{2} \sin ^2 2 x \\ & \text { Period of } \cos ^4 x+\sin ^4 x \\ & =\text { Period of } 1-\frac{1}{2} \sin ^2 2 x \\ & \therefore \quad \gamma=\frac{\pi}{2} \\ & \frac{\alpha \gamma}{\beta}=\frac{12 \pi}{4 \pi \times 2}=\frac{3}{2} \end{aligned} $

Given, $\tan \theta=-\frac{3}{4}$

If $\theta$ does not lie in second quadrant, then $\theta$ lies in fourth quadrant as $\tan \theta$ is negative only in second and fourth

$ \begin{aligned} & \theta \in\left(\frac{3 \pi}{2}, 2 \pi\right) \Rightarrow \frac{\theta}{2} \in\left(\frac{3 \pi}{4}, \pi\right) \\ \Rightarrow & 2 \theta \in(3 \pi, 4 \pi) \end{aligned} $

$\Rightarrow \theta / 2$ lies in second quadrant.

$2 \theta$ lies in third and fourth quadrant.

$ \begin{aligned} \tan \theta & =-\frac{3}{4} \\ \Rightarrow & \frac{2 \tan \frac{\theta}{2}}{1-\tan ^2 \frac{\theta}{2}}=-\frac{3}{4} \\ \Rightarrow & 8 \tan \frac{\theta}{2}=-3+3 \tan ^2 \frac{\theta}{2} \\ \Rightarrow & 3 \tan ^2 \frac{\theta}{2}-8 \tan \frac{\theta}{2}-3=0 \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad 3 \tan ^2 \frac{\theta}{2}-9 \tan \frac{\theta}{2}+\tan \frac{\theta}{2}-3=0 \\ & \Rightarrow \quad \tan \frac{\theta}{2}=-\frac{1}{3} \quad\left[\because \tan \frac{\theta}{2} \neq 3\right] \end{aligned} $

$ \begin{aligned} \sin 2 \theta & =\frac{2 \tan \theta}{1+\tan ^2 \theta} \\ & =\frac{2\left(-\frac{3}{4}\right)}{1+\frac{9}{16}}=-\frac{3}{2} \times \frac{16}{25} \\ & =-\frac{24}{25} \\ \therefore \tan \frac{\theta}{2}+\sin 2 \theta & =-\frac{1}{3}-\frac{24}{25} \\ & =\frac{-25-72}{75}=-\frac{97}{75} \end{aligned} $

$ \begin{aligned} &\text {We have to find, }\\ &\begin{array}{r} \cos ^2 76^{\circ}+\sin ^2 46^{\circ}+\sin 76^{\circ} \cos 46^{\circ} \\ =\cos ^2 76^{\circ}+\sin ^2\left(76^{\circ}-30^{\circ}\right) \\ +\sin 76^{\circ} \cos \left(76^{\circ}-30^{\circ}\right) \\ =\cos ^2 76^{\circ}+\left[\sin 76^{\circ} \cos 30^{\circ}-\cos 76^{\circ}\right. \\ \left.\sin 30^{\circ}\right]^2+\sin 76^{\circ}\left[\cos 76^{\circ} \cos 30^{\circ}\right. \\ \left.+\sin 76^{\circ} \sin 30^{\circ}\right] \end{array} \end{aligned} $

$ \begin{aligned} & =\cos ^2 76+\left(\frac{\sqrt{3} \sin 76^{\circ}-\cos 76^{\circ}}{2}\right)^2 \\ & +\sin 76^{\circ}\left(\frac{\sqrt{3} \cos 76^{\circ}}{2}+\frac{\sin 76^{\circ}}{2}\right) \\ & \begin{aligned} & 3 \sin ^2 76^{\circ}+\cos ^2 76^{\circ} \\ & \cos ^2 76^{\circ}+\frac{-2 \sqrt{3} \sin 76^{\circ} \cos 76^{\circ}}{4} \end{aligned} \\ & +\frac{\sqrt{3} \cos 76^{\circ} \sin 76^{\circ}+\sin ^2 76^{\circ}}{2} \\ & =\frac{1}{4}\left[4 \cos ^2 76^{\circ}+3 \sin ^2 76^{\circ}+\cos ^2 76^{\circ}\right.-2 \sqrt{3} \sin 76^{\circ} \cos 76^{\circ} \\ & \left.+2 \sqrt{3} \sin 76^{\circ} \cos 76^{\circ}+2 \sin ^2 76^{\circ}\right] \end{aligned} $

$ \begin{aligned} & =\frac{1}{4}\left[5 \cos ^2 76^{\circ}+5 \sin ^2 76^{\circ}\right] \\ & =\frac{5}{4} \quad\left[\because \cos ^2 76^{\circ}+\sin ^2 76^{\circ}=1\right] \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { }|\sin \alpha-\cos \alpha|=3 / 4 \\ & \Rightarrow \quad(\sin \alpha-\cos \alpha)^2=9 / 16 \\ & \Rightarrow \quad \sin ^2 \alpha+\cos ^2 \alpha-2 \sin \alpha \cos \alpha=\frac{9}{16} \\ & \Rightarrow \quad 1-\frac{9}{16}=2 \sin \alpha \cos \alpha \\ & \Rightarrow \quad 2 \sin \alpha \cos \alpha=\frac{7}{16} \\ & \begin{aligned} \mid \sin \alpha & +\cos \alpha \mid \\ & =\sqrt{\sin ^2 \alpha+\cos ^2 \alpha+2 \sin \alpha \cos \alpha} \\ & =\sqrt{1+\frac{7}{16}}=\sqrt{\frac{23}{16}}=\frac{\sqrt{23}}{4} \\ & \begin{aligned} \mid \sec 2 \alpha & -\tan 2 \alpha \mid \\ & =\left|\frac{1+\tan ^2 \alpha}{1-\tan ^2 \alpha}-\frac{2 \tan \alpha}{1-\tan ^2 \alpha}\right| \\ & =\left|\frac{\left(1-\tan ^2\right)^2}{1-\tan ^2 \alpha}\right|=\left|\frac{1-\tan \alpha}{1+\tan \alpha}\right| \\ & =\frac{(\cos \alpha-\sin \alpha)}{(\cos \alpha+\sin \alpha)}=\frac{3}{4} \div \frac{\sqrt{23}}{4}=\frac{3}{\sqrt{23}} \end{aligned} \end{aligned} \text { } \end{aligned} \end{aligned} $

Given,

$ \begin{aligned} & \frac{1}{\sin 45^{\circ} \cdot \sin 46^{\circ}}+\frac{1}{\sin 46^{\circ} \sin 47^{\circ}}+\ldots \\ & \text { upto } 45 \text { terms }=\frac{1}{\sin x^{\circ}} \\ & \Rightarrow \frac{1}{\sin 1^{\circ}}\left[\frac{\sin 1^{\circ}}{\sin 45^{\circ} \sin 46^{\circ}}+\frac{\sin 1^{\circ}}{\sin 46^{\circ} \sin 47^{\circ}}\right. \\ & +\ldots \text { upto } 45 \text { terms }] \\ & =\frac{1}{\sin x} \end{aligned} $

[ ∵ multiplying and dividing by $\sin 1^{\circ}$ ]

$ \begin{aligned} & \Rightarrow \frac{1}{\sin 1^{\circ}}\left[\frac{\sin \left(46^{\circ}-45^{\circ}\right)}{\sin 45^{\circ} \sin 46^{\circ}}+\frac{\sin \left(47^{\circ}-46^{\circ}\right)}{\sin 46^{\circ} \sin 47^{\circ}}\right. \\ & +\ldots \text { upto } 45 \text { terms }] \frac{1}{\sin x} \\ & \Rightarrow \frac{1}{\sin 1^{\circ}}\left[\cot 45^{\circ}-\cot 46^{\circ}+\cot 46^{\circ}\right. \\ & \left.-\cot 47^{\circ}+\ldots+\cot 89^{\circ}-\cot 90^{\circ}\right]=\frac{1}{\sin x^{\circ}} \\ & \Rightarrow \quad \frac{1}{\sin 1^{\circ}}\left[\cot 45^{\circ}-\cot 90^{\circ}\right]=\frac{1}{\sin x^{\circ}} \\ & \Rightarrow \quad x=1 \\ & \Rightarrow \quad \sin \left(\frac{\pi}{2} x\right)=\sin \frac{\pi}{2}=1 \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { } \sinh x=\frac{-1}{2} \\ & \Rightarrow \quad \frac{e^x-e^{-x}}{2}=-\frac{1}{2} \Rightarrow e^x-\frac{1}{e^x}=-1 \\ & \Rightarrow \quad\left(e^x\right)^2+e^x-1=0 \\ & \text { Let } e^x=y \\ & \Rightarrow \quad y^2+y-1=0 \\ & \Rightarrow \quad y=\frac{-1 \pm \sqrt{1+4}}{2}=\frac{-1+\sqrt{5}}{2} \\ & \Rightarrow \quad e^x=\frac{-1+\sqrt{5}}{2} \\ & \Rightarrow e^{-x}=\frac{2}{-1+\sqrt{5}} \times \frac{(1+\sqrt{5})}{(1+\sqrt{5})} \\ & \quad=\frac{2(\sqrt{5}+1)}{4}=\frac{\sqrt{5}+1}{2} \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { }\\ &\begin{aligned} So, \tanh & 2 x=\frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}} \\ & =\frac{\left(\frac{\sqrt{5}-1}{2}\right)^2-\left(\frac{5+1}{2}\right)^2}{\left(\frac{\sqrt{5}-1}{2}\right)^2+\left(\frac{\sqrt{5}+1}{2}\right)^2} \\ & =\frac{5+1-2 \sqrt{5}-5-1-2 \sqrt{5}}{5+1-2 \sqrt{5}+5+1+2 \sqrt{5}} \\ & =\frac{-4 \sqrt{5}}{12}=\frac{-\sqrt{5}}{3} \end{aligned} \end{aligned} $

$\cos x+\cos y=p\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

$ \sin x+\sin y=q \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

On adding Eqs. (i) and (ii) after doing square

$ \begin{aligned} & \left(\cos ^2 x+\cos ^2 y+2 \cos x \cdot \cos y\right) \\ & +\left(\sin ^2 x+\sin ^2 y+2 \sin x \cdot \sin y\right)=p^2+q^2 \end{aligned} $

$ \begin{array}{rrl} \Rightarrow & 2+2 \cos (x-y) & =p^2+q^2 \\ \Rightarrow & 2[1+\cos (x-y)] & =p^2+q^2 \\ \Rightarrow & 2 \cdot 2 \cos ^2\left(\frac{x-y}{2}\right) & =p^2+q^2 \\ \Rightarrow & \cos ^2\left(\frac{x-y}{2}\right) & =\frac{p^2+q^2}{4} \\ \therefore & \cos \left(\frac{x-y}{2}\right) & = \pm \frac{\sqrt{p^2+q^2}}{2} \end{array} $

$ \text { } \begin{aligned} & \text { } A+B+C=\frac{3 \pi}{2} \\ & \cos 2 A+\cos 2 B+\cos 2 C \\ = & 2 \cos \left(\frac{2 A+2 B}{2}\right) \cdot \cos \left(\frac{2 A-2 B}{2}\right)+\cos 2 C \\ = & 2 \cos \left(\frac{3 \pi}{2}-C\right) \cdot \cos (A-B)+\cos 2 C \\ = & 2(-\sin C) \cdot \cos (A-B)+\left(-2 \sin ^2 C+1\right) \\ = & -2 \sin C[\cos (A-B)+\sin C]+1 \\ = & -2 \sin C[\cos (A-B) \left.+\sin \left(\frac{3 \pi}{2}-(A+B)\right)\right]+1 \\ = & -2 \sin C[\cos (A-B)-\cos (A+B)]+1 \\ = & -2 \sin C(2 \sin A \cdot \sin B)+1 \end{aligned} $

$ \begin{aligned} & =-4 \sin A \cdot \sin B \cdot \sin C+1 \\ & \begin{array}{r} \Rightarrow 4 \sin A \cdot \sin B \cdot \sin C+(\cos 2 A \\ \quad+\cos 2 B+\cos 2 C)=1 \end{array} \\ & =-\sin (A+B+C) \\ & \quad\left[\because \sin (A+B+C)=\sin \frac{3 \pi}{2}=-1\right] \end{aligned} $

$ \text { } \begin{aligned} & \frac{e^{4 x}+e^{-4 x}+14}{4\left(e^x-e^{-x}\right)^2}=\frac{\left(e^{2 x}-e^{-2 x}\right)^2}{4\left(e^x-e^{-x}\right)^2}+16 \\ = & \left(\frac{e^x+e^{-x}}{2}\right)^2+\frac{4}{\left(e^x-e^{-x}\right)^2} \\ = & \left(\frac{e^x+e^{-x}}{2}\right)^2+\frac{\left(e^x+e^{-x}\right)^2-\left(e^x-e^{-x}\right)^2}{\left(e^x-e^{-x}\right)^2} \\ = & \frac{\left(e^x+e^{-x}\right)^2}{4}-1+\frac{\left(e^x+e^{-x}\right)^2}{\left(e^x-e^{-x}\right)^2} \\ = & \frac{\left(e^x-e^{-x}\right)^2}{4}+\frac{\left(e^x+e^{-x}\right)^2}{\left(e^x-e^{-x}\right)^2} \\ = & \left(\frac{e^x-e^{-x}}{2}\right)^2+\left(\frac{e^x+e^{-x}}{e^x-e^{-x}}\right)^2 \\ = & \sinh ^2 x+\operatorname{coth}^2 x \end{aligned} $

$ \begin{aligned} &\text { } \tanh x=1 / 2\\ &\begin{aligned} & \operatorname{sech} 2 x=\frac{1-\tan ^2 h x}{1+\tan ^2 h x}=\frac{1-\frac{1}{4}}{1+\frac{1}{4}}=\frac{3}{5} \\ & \sinh 2 x=\frac{2 \tanh x}{1-\tan ^2 h x}=\frac{2 \times \frac{1}{2}}{1-\frac{1}{4}}=\frac{4}{3} \\ & \sinh 2 x-\operatorname{sech} 2 x=\frac{4}{3}-\frac{3}{5}=\frac{11}{15} \end{aligned} \end{aligned} $

$A>B$ and $A, B$ are acute angles.

Given, $\sin (A-B)=\frac{16}{65}$ and $\sin B=\frac{5}{13}$ Then, $\cos (A-B)=\frac{63}{65}$ and $\cos B=\frac{12}{13}$

$ \begin{aligned} & \sin A=\sin ((A-B)+B) \\ & =\sin (A-B) \cos B+\cos (A-B) \sin B \\ & =\frac{16}{65} \times \frac{12}{13}+\frac{63}{65} \times \frac{5}{13}=\frac{507}{845}=\frac{3}{5} \\ & \cos A=\cos ((A-B)+B) \\ & =\cos (A-B) \cos B-\sin (A-B) \sin B \end{aligned} $

$ \begin{aligned} & =\frac{63}{65} \times \frac{12}{13}-\frac{16}{65} \times \frac{5}{13}=\frac{676}{845}=\frac{4}{5} \\ \therefore \quad & \tan A+\cot A=\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A} \\ = & \frac{1}{\sin A \cos A}=\frac{1}{\frac{3}{5} \times \frac{4}{5}}=\frac{25}{12} \end{aligned} $

$ \begin{aligned} &\text { Given, } \tan A=2 / 3\\ &\begin{aligned} & \because \quad \sin 4 A=2 \sin 2 A \cdot \cos 2 A \\ & \quad=2 \cdot \frac{2 \tan A}{1+\tan ^2 A} \times \frac{1-\tan ^2 A}{1+\tan ^2 A} \\ & \quad=4 \times \frac{\frac{2}{3}}{1+\frac{4}{9}} \times \frac{1-\frac{4}{9}}{1+\frac{4}{9}}=\frac{120}{169} \end{aligned} \end{aligned} $

$ \text { Given, } \frac{\sqrt{2} \cos 45^{\circ}+\cos 56^{\circ}+\cos 58^{\circ}-\cos 66^{\circ}}{\sqrt{2} \cos 28^{\circ} \cos 29^{\circ} \sin 33^{\circ}} $

$ \begin{aligned} & =\frac{\sqrt{2}\left(\frac{1}{\sqrt{2}}\right)+\cos 56^{\circ}+\cos 58^{\circ}-\cos 66^{\circ}}{\sqrt{2} \cos 28^{\circ} \cos 29^{\circ} \sin 33^{\circ}} \\ & =\frac{\left(1-\cos 66^{\circ}\right)+2 \cos 57^{\circ} \cos 1^{\circ}}{\sqrt{2} \cos 28^{\circ} \cos 29^{\circ} \sin 33^{\circ}} \\ & =\frac{2 \sin ^2 33^{\circ}+2 \sin 33^{\circ} \cos 1^{\circ}}{\sqrt{2} \cos 28^{\circ} \cos 29^{\circ} \sin 33^{\circ}} \\ & =2 \sin 33^{\circ}\left(\frac{\sin 33^{\circ}+\cos 1^{\circ}}{\sqrt{2} \cos 28^{\circ} \cos 29^{\circ} \sin 33^{\circ}}\right) \\ & =\sqrt{2}\left(\frac{\cos 57^{\circ}+\cos 1^{\circ}}{\cos 28^{\circ} \cos 29^{\circ}}\right) \\ & =\sqrt{2}\left(\frac{2 \cos 29^{\circ} \cos 28^{\circ}}{\cos 28^{\circ} \cos 29^{\circ}}\right)=2 \sqrt{2} \end{aligned} $

For $\theta=\pi / 12$

$ \begin{aligned} & x=\log \left(\cot \left(\frac{\pi}{4}+\frac{\pi}{12}\right)\right) \\ & x=\log \left(\frac{1}{\sqrt{3}}\right) \end{aligned} $

$ \begin{aligned} Now,\cosh x & =\frac{e^x+e^{-x}}{2} \\ & =\frac{e^{\log \frac{1}{\sqrt{3}}}+e^{-\log \frac{1}{\sqrt{3}}}}{2} \\ & =\frac{\frac{1}{\sqrt{3}}+\sqrt{3}}{2}=\frac{2}{\sqrt{3}} \end{aligned} $

$ \begin{aligned} &\text { We know that, }\\ &\begin{aligned} & \sinh x=\frac{e^x-e^{-x}}{2} \text { and } \cosh x=\frac{e^x+e^{-x}}{2} \\ & \because 2 \cosh (x+y) \cdot \sinh (x-y)+\sinh 2 y \\ & =2 \cdot \frac{e^{x+y}+e^{-(x+y)}}{2} \cdot \frac{e^{(x-y)}-e^{-(x-y)}}{2} +\frac{e^{2 y}-e^{-2 y}}{2} \end{aligned} \end{aligned} $

$ \begin{aligned} & =\frac{e^{2 x}-e^{2 y}+e^{-2 y}-e^{-2 x}+e^{2 y}-e^{-2 y}}{2} \\ & =\frac{e^{2 x}-e^{-2 x}}{2}=\sinh 2 x \end{aligned} $

$\begin{aligned} \text{Given,}\quad & \sin ^4 \theta \cos ^2 \theta=\sum_{n=0}^{\infty} a_{2 n} \cos 2 n \theta \\ \Rightarrow & \left(\sin ^2 \theta\right)^2 \cos ^2 \theta=\sum_{n=0}^{\infty} a_{2 n} \cos 2 n \theta \\ \Rightarrow & \left(1-\cos ^2 \theta\right)^2 \cos ^2 \theta=\sum_{n=0}^{\infty} a_{2 n} \cos 2 n \theta \\ \Rightarrow & \left(1+\cos ^4 \theta-2 \cos ^2 \theta\right) \cos ^2 \theta=\sum_{n=0}^{\infty} a_{2 n} \cos 2 n \theta \\ \therefore & \cos ^2 \theta+\cos ^4 \theta \cos ^2 \theta-2 \cos ^4 \theta \\ & =\sum_{n=0}^{\infty} a_{2 n} \cos 2 n \theta \\ \Rightarrow & \left(\frac{\cos 2 \theta+1}{2}\right)+\left(\frac{\cos 2 \theta+1}{2}\right)^2\left(\frac{1+\cos 2 \theta}{2}\right) \\ & -2\left(\frac{\cos 2 \theta+1}{2}\right)^2=\sum_{n=0}^{\infty} a_{2 n} \cos 2 n \theta \end{aligned}$

On comparing both sides, we get minimum value of n is 1.

$\begin{aligned} \sin \theta & =\frac{-3}{4} \\ \cos \theta & =\sqrt{1-\sin ^2 \theta} \\ \cos \theta & =\sqrt{1-\left(\frac{-3}{4}\right)^2} \\ \cos \theta & =\sqrt{\frac{16-9}{4}}=\sqrt{\frac{7}{16}} \\ \therefore \quad \sin 2 \theta & =2 \sin \theta \cdot \cos \theta \\ & =2 \times\left(\frac{-3}{4}\right)\left(\frac{\sqrt{7}}{4}\right)=\frac{-3 \sqrt{7}}{8} \end{aligned}$

Here,

$\begin{aligned} & \frac{1}{\sin 1^{\circ} \sin 2^{\circ}}+\frac{1}{\sin 2^{\circ} \sin 3^{\circ}}+\ldots+\frac{1}{\sin 89^{\circ} \sin 90^{\circ}} \\ & \quad=\frac{1}{\sin 1^{\circ}}\left[\begin{array}{l} \frac{\sin \left(2^{\circ}-1^{\circ}\right)}{\sin 1^{\circ} \sin 2^{\circ}}+\frac{\sin \left(3^{\circ}-2^{\circ}\right)}{\sin 2^{\circ} \sin 3^{\circ}}+\ldots . . \\ +\frac{\sin \left(90^{\circ}-89^{\circ}\right)}{\sin 89^{\circ} \sin 90^{\circ}} \end{array}\right] \end{aligned}$

We know that

$\sin (A-B)=\sin A \cdot \cos B-\cos A \cdot \sin B$

$\begin{array}{r} =\frac{1}{\sin 1^{\circ}}\left[\begin{array}{l} \frac{\sin 2^{\circ} \cdot \cos 1^{\circ}-\cos 2^{\circ} \cdot \sin 1^{\circ}}{\sin 1^{\circ} \cdot \sin 2^{\circ}} \\ +\frac{\sin 3^{\circ} \cdot \cos 2^{\circ}-\cos 3^{\circ} \cdot \sin 2^{\circ}}{\sin 3^{\circ} \cdot \sin 2^{\circ}}+\ldots . \\ \ldots .+\frac{\sin 90^{\circ} \cdot \cos 89^{\circ}-\cos .90^{\circ} \cdot \sin 89^{\circ}}{\sin 89^{\circ} \cdot \sin 90^{\circ}} \end{array}\right] \end{array}$

$\begin{aligned} & =\frac{1}{\sin 1^{\circ}}\left[\cot 1^{\circ}-\cot 2^{\circ}+\cot 2^{\circ}-\cot 3^{\circ}+\ldots .\right. \\ & \left.+\quad+\cot 89^{\circ}-\cot 90^{\circ}\right] \\ & =\frac{1}{\sin 1^{\circ}}\left[\cot 1^{\circ}-\cot 90^{\circ}\right]=\frac{\cot 1^{\circ}}{\sin 1^{\circ}}=\frac{\cos 1^{\circ}}{\sin ^2 1^{\circ}} \end{aligned}$

For option I,

$\begin{aligned} \sin \left(-292^{\circ}\right) & =-\sin 292^{\circ} \\ & =-\sin \left(360^{\circ}-68^{\circ}\right)=-\sin 68^{\circ} \\ & =-(-)=+ \text { ve value } \end{aligned}$

So, clearly $\sin \left(-292^{\circ}\right)$ give +ve value.

For option II,

$\begin{aligned} \tan \left(-193^{\circ}\right) & =-\tan 193^{\circ} \\ & =-\tan \left(180^{\circ}+13^{\circ}\right) \\ & =-\tan 13^{\circ} \text { (-ve value) } \end{aligned}$

For option III,

$\begin{aligned} \cos \left(-207^{\circ}\right) & =\cos 207^{\circ} \\ & =\cos \left(180+27^{\circ}\right) \\ & =-\cos 27^{\circ}(- \text { ve value }) \end{aligned}$

For option IV,

$\begin{aligned} \cot \left(-222^{\circ}\right) & =\cot \left(180^{\circ}+42^{\circ}\right) \\ & =+\cot 42^{\circ}(+ \text { ve value }) \end{aligned}$

Here,

$\begin{aligned} & \sin ^2\left(\frac{2 \pi}{3}\right)+\cos ^2\left(\frac{5 \pi}{6}\right)-\tan ^2\left(\frac{3 \pi}{4}\right) \\ = & \sin ^2\left(\pi-\frac{\pi}{3}\right)+\cos ^2\left(\pi-\frac{\pi}{6}\right)-\tan ^2\left(\pi-\frac{\pi}{4}\right) \\ = & \sin ^2\left(\frac{\pi}{3}\right)+\left(-\cos \left(\frac{\pi}{6}\right)\right)^2-\left(-\tan \frac{\pi}{4}\right)^2 \end{aligned}$

$\left[\begin{array}{l} \therefore \sin (\pi-x)=\sin x \\ \cos (\pi-x)=-\cos x \\ \tan (\pi-x)=-\tan x \end{array}\right]$

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

$\begin{aligned} & \sin 5 \theta=\sin (3 \theta+2 \theta) \\ & =\sin 3 \theta \cos 2 \theta+\cos 3 \theta \cdot \sin 2 \theta \\ & =\left(3 \sin \theta-4 \sin ^3 \theta\right) \cdot\left(1-2 \sin ^2 \theta\right) \\ & \quad \quad+\left(4 \cos ^3 \theta-3 \cos \theta\right) \cdot(2 \sin \theta \cos \theta) \\ & =3 \sin \theta-6 \sin ^3 \theta-4 \sin ^3 \theta+8 \sin ^5 \theta+8 \sin \theta \cos ^4 \theta \\ & \quad-6 \sin \theta \cos ^2 \theta \\ & =3 \sin \theta-10 \sin ^3 \theta+8 \sin ^5 \theta+8 \sin \theta \cos ^4 \theta \\ & -6 \sin \theta \cos ^2 \theta \end{aligned}$

$\begin{aligned} & =3 \sin \theta-10 \sin ^3 \theta+8 \sin ^5 \theta+8 \sin \theta\left(1-\sin ^2 \theta\right)^2 \\ & -6 \sin \theta\left(1-\sin ^2 \theta\right) \\ & =3 \sin \theta-10 \sin ^3 \theta+8 \sin ^5 \theta+8 \sin \theta\left(1+\sin ^4 \theta-2 \sin ^2 \theta\right) \\ & -6 \sin \theta+6 \sin ^3 \theta \\ & =-3 \sin \theta-4 \sin ^3 \theta+8 \sin ^5 \theta+8 \sin \theta \\ & +8 \sin ^5 \theta-16 \sin ^3 \theta \\ & =5 \sin \theta-20 \sin ^3 \theta+16 \sin ^5 \theta \\ & =5 \sin \theta-20 \sin \theta\left(1-\cos ^2 \theta\right)+16\left(1-\cos ^2 \theta\right)^2 \sin \theta \\ & =5 \sin \theta-20 \sin \theta+20 \sin \theta \cos ^2 \theta+16 \sin \theta \\ & \left(1+\cos ^4 \theta-2 \cos ^2 \theta\right) \\ & =-15 \sin \theta+20 \sin \theta \cos ^2 \theta+16 \sin \theta+16 \cos ^4 \theta \sin \theta \\ & -32 \cos ^2 \theta \sin \theta \\ & =16 \cos ^4 \theta \sin \theta-12 \cos ^2 \theta \sin \theta+\sin \theta \\ & \end{aligned}$

Given $A+B+C=\pi$

$\begin{aligned} & \therefore \cos B=\cos A \cdot \cos C \\ & \Rightarrow \quad \cos \{\pi-(A+C)\}=\cos A \cdot \cos C \\ & \Rightarrow \quad-\cos (A+C)=\cos A \cdot \cos C \\ & \Rightarrow \quad-\cos A \cdot \cos C+\sin A \cdot \sin C=\cos A \cdot \cos C \\ & \Rightarrow \quad \sin A \cdot \sin C=2 \cos A \cos C \\ & \Rightarrow \quad \frac{\sin A}{\cos A} \cdot \frac{\sin C}{\cos C}=2 \\ & \Rightarrow \quad \tan A \cdot \tan C=2 \end{aligned}$

Observe that $\begin{aligned} \tan \left(\frac{7 \pi}{8}\right) & =\tan \left(\pi-\frac{\pi}{8}\right) \\ & =-\tan \left(\frac{\pi}{8}\right)\end{aligned}$

Now, using $\tan (2 \theta)=\frac{2 \tan \theta}{1-\tan ^2 \theta}$

Let $\theta=\frac{\pi}{8}$

So, $\tan \left(2 \cdot \frac{\pi}{8}\right)=\frac{2 \tan \frac{\pi}{8}}{1-\tan ^2 \frac{\pi}{8}}$

$\Rightarrow \tan \frac{\pi}{4}=\frac{2 \tan \frac{\pi}{8}}{1-\tan ^2 \frac{\pi}{8}}$

On putting $\tan \frac{\pi}{8}=x$,

$\begin{aligned} 1 & =\frac{2 x}{1-x^2} \\ \Rightarrow \quad & \\ \Rightarrow \quad 1-x^2-2 x & =0 \Rightarrow x^2+2 x-1=0 \\ x & =-1-\sqrt{2}, \sqrt{2}-1 \end{aligned}$

But, $\tan \frac{\pi}{8}>0$ so, only take $\tan \frac{\pi}{8}=\sqrt{2}-1$

Thus, $\tan \left(\frac{7 \pi}{8}\right)=-\tan \left(\frac{\pi}{8}\right)=-(\sqrt{2}-1)=1-\sqrt{2}$

Given,

$\begin{aligned} & 1+\sec ^2 x \sin ^2 x \\ &=1+\left(\frac{1}{\cos ^2 x}\right) \sin ^2 x \\ &=1+\tan ^2 x=\sec ^2 x \end{aligned}$

$\begin{aligned} & \text { } \cos ^4 \theta=a \cos 4 \theta+b \cos 2 \theta+c \\ & \Rightarrow \cos ^4 \theta=a\left[2 \cos ^2 2 \theta-1\right]+b\left[2 \cos ^2 \theta-1\right]+c \\ & \Rightarrow \cos ^4 \theta=a \cdot 2 \cos ^2 2 \theta-a+2 b \cos ^2 \theta-b+c \\ & \Rightarrow \cos ^4 \theta=2 a(\cos 2 \theta)^2+2 b \cos ^2 \theta-a-b+c \\ & \Rightarrow \cos ^4 \theta=2 a\left(2 \cos ^2 \theta-1\right)^2+2 b \cos ^2 \theta-a-b+c \\ & \Rightarrow \cos ^4 \theta=2 a\left(4 \cos ^4 \theta+1-4 \cos ^2 \theta\right) \\ & +2 b \cos ^2 \theta-a-b+c \\ & \Rightarrow \cos ^4 \theta=8 a \cos ^4 \theta+2 a-8 a \cos ^2 \theta \\ & +2 b \cos ^2 \theta-a-b+c \\ & \Rightarrow \cos ^4 \theta=8 a \cos ^4 \theta+\cos ^2 \theta(-8 a+2 b)+a-b+c \\ \end{aligned}$

On comparing the corresponding coefficients, we get

$\begin{aligned} & 8 a=1,-8 a+2 b=0 \text { and } a-b+c=0 \\ & \Rightarrow \quad a=\frac{1}{8} \\ & \therefore \quad-8\left(\frac{1}{8}\right)+2 b=0 \text { gives } b=\frac{1}{2} \\ & \text { and } \quad \frac{1}{8}-\frac{1}{2}+c=0 \text { gives } c=\frac{3}{8} \text {. } \\ \end{aligned}$

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

$(1+\tan 1^{\circ})(1+\tan 2^{\circ})(1+\tan 3^{\circ}) \ldots (1+\tan 45^{\circ})$

Considering the following two terms

$(1+\tan x)\left(1+\tan \left(45^{\circ}-x\right)\right)$

$\begin{aligned} & =1+\tan x+\tan \left(45^{\circ}-x\right)+\tan x \cdot \tan \left(45^{\circ}-x\right) \\ & =1+\left[\tan x+\tan \left(45^{\circ}-x\right)\right]+\tan x \cdot \tan \left(45^{\circ}-x\right) \\ & =1+\tan \left(x+\left(45^{\circ}-x\right)\right) \cdot\left[1-\tan x \cdot \tan \left(45^{\circ}-x\right)\right] \\ & \quad+\tan x \cdot \tan \left(45^{\circ}-x\right) \\ & =1+\left[1-\tan x \cdot \tan \left(45^{\circ}-x\right)\right]+\tan x \cdot \tan \left(45^{\circ}-x\right) \\ & =1+1=2 \end{aligned}$

There are 22 such terms in the given multiplication

$\begin{aligned} & =(2)^{22} \cdot\left(1+\tan 45^{\circ}\right) \\ & =2^{22} \cdot 2^1=2^{23}=2^n \\ & \therefore \quad n=23 \end{aligned}$

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

$\operatorname{cosech} x=\frac{4}{5}$

We know that, $\sinh x=\frac{1}{\operatorname{cosech} x}$

$\therefore \sinh x=\frac{5}{4}$

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

Then, $|\tan\theta|=5/12$

and $ - \tan \theta = - \left( {{{ - 5} \over {12}}} \right) = {5 \over {12}}$

$\therefore$ $|\tan \theta | = - \tan \theta $

$\begin{aligned} & \text { } \cos \frac{\pi}{4} \cos \frac{\pi}{8} \cos \frac{\pi}{16} \cos \frac{\pi}{32} \\ & =\cos \frac{\pi}{2^2} \cdot \cos \frac{\pi}{2^3} \cdot \cos \frac{\pi}{2^4} \cdot \cos \frac{\pi}{2^5} \\ & =\frac{1}{2 \sin \left(\frac{\pi}{2^5}\right)} \\ & {\left[2 \sin \left(\frac{\pi}{2^5}\right) \cdot \cos \left(\frac{\pi}{2^5}\right) \cdot \cos \frac{\pi}{2^2} \cdot \cos \frac{\pi}{2^3} \cdot \cos \frac{\pi}{2^4}\right]} \\ & =\frac{1}{2} \operatorname{cosec}\left(\frac{\pi}{2^5}\right)\left[\sin \frac{\pi}{2^4} \cdot \cos \frac{\pi}{2^4} \cdot \cos \frac{\pi}{2^2} \cdot \cos \frac{\pi}{2^3}\right] \\ & =\frac{1}{4} \operatorname{cosec}\left(\frac{\pi}{2^5}\right)\left[\sin \frac{\pi}{2^3} \cdot \cos \frac{\pi}{2^3} \cdot \cos \frac{\pi}{2^2}\right] \\ & =\frac{1}{8} \operatorname{cosec}\left(\frac{\pi}{2^5}\right)\left[\sin \frac{\pi}{2^2} \cdot \cos \frac{\pi}{2^2}\right] \\ & =\frac{1}{16} \operatorname{cosec} \frac{\pi}{2^5} \sin \frac{\pi}{2} \\ & =2^{-4} \operatorname{cosec} \frac{\pi}{2^5} \quad\left[\because \sin \frac{\pi}{2}=1\right] \\ & \therefore m=-4 \text { and } n=2^5 \text {, then } \\ & m+n=-4+32=28 \\ & \end{aligned}$

$\begin{aligned} & A+B+C=\frac{3 \pi}{2} \\ & A+B=\frac{3 \pi}{2}-C \\ & \cos (A+B)=\cos \left(\frac{3 \pi}{2}-C\right)=-\sin C \end{aligned}$

Now, $\cos 2 A+\cos 2 B+\cos 2 C$

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