Trigonometric Ratios & Identities

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} $

Given,

$ \begin{aligned} & A=\left[\begin{array}{ccc} \cos ^2 37 & \cos ^2 53 & \cot 135 \\ \sin ^2 76 & \sin 270 & \sin ^2 14 \\ \cos 180 & \cos ^2 28 & \cos ^2 62 \end{array}\right] \\ & \therefore A=\left[\begin{array}{ccc} \cos ^2 37 & \cos ^2 53 & -1 \\ \sin ^2 76 & -1 & \sin ^2 14 \\ -1 & \cos ^2 28 & \cos ^2 62 \end{array}\right] \\ & \therefore|A|=\left|\begin{array}{ccc} \cos ^2 37 & \cos ^2 53 & -1 \\ \sin ^2 76 & -1 & \sin ^2 14 \\ -1 & \cos ^2 28 & \cos ^2 62 \end{array}\right| \end{aligned} $

Use $C_1 \rightarrow C_1+C_2$ and $C_2 \longrightarrow C_2+C_3$

$ |A|=\left\lvert\, \begin{array}{cc} \cos ^2 37+\cos ^2 53 & \cos ^2 53-1 \\ \sin ^2 76-1 & -1+\sin ^2 14 \\ -1+\cos ^2 28 & \cos ^2 28+\cos ^2 62 \end{array}\right. $

$\left. \matrix{ - 1\,\,\,\,\,\,\,\,\,\,\, \hfill \cr {\sin ^2}14\,\,\, \hfill \cr {\cos ^2}62\, \hfill \cr} \right|$$ \begin{aligned} & |A|=\left|\begin{array}{ccc} 1 & -\sin ^2 53 & -1 \\ -\cos ^2 76 & -\cos ^2 14 & \sin ^2 14 \\ -\sin ^2 28 & 1 & \cos ^2 62 \end{array}\right| \\ & |A|=-\left|\begin{array}{ccc} 1 & -\sin ^2 53 & +1 \\ -\sin ^2 14 & -\cos ^2 14 & -\sin ^2 14 \\ -\cos ^2 62 & 1 & -\cos ^2 62 \end{array}\right| \\ & \Rightarrow|A|=0 \\ & \text { Now 3-|A|=3 } \\ & \therefore \quad \text { (A) → (iii) } \end{aligned} $

(B) Given $\frac{\cos (6 x-4)+\sec (3-4 x)}{\cot (5 x+3)+\sin (3 x+4)}$

∴ Period of $\cos (6 x-4)=\frac{2 \pi}{6}=\frac{\pi}{3}$

Period of $\sec (3-4 x)=\frac{2 \pi}{4}=\frac{\pi}{2}$

Period of $\cot (5 x+3)=\frac{\pi}{5}$

Period of $\sin (3 x+4)=\frac{2 \pi}{3}$

Now required period = lim of $\left(\frac{\pi}{3}, \frac{\pi}{2}, \frac{\pi}{5}, \frac{2 \pi}{3}\right)=2 \pi$

Now, given $\frac{2 k \pi}{5}=2 \pi$

$ \Rightarrow k=5 $

(B) $\rightarrow(\mathrm{V})$

(C) given

$ \begin{aligned} & y=\cos ^2\left(\frac{\pi}{4}-x\right)+(\sin x-\cos x)^2 \\ & \therefore y=\frac{1}{2}(\cos x+\sin x)^2+(\sin x-\cos x)^2 \\ & \quad y=\frac{1}{2}(1+\sin 2 x)+(1-\sin 2 x) \\ & \Rightarrow \quad y=\frac{3}{2}-\frac{\sin 2 x}{2} \end{aligned} $

$\therefore y_{\text {max }}$ when $\sin 2 x=-1$

$ \therefore \quad y_{\max }=\frac{3}{2}+\frac{1}{2}=2 $

(C) → (ii)

(D) It $x+y+z=0^{\circ}$

$ \begin{array}{r} \Rightarrow \sin 2 x+\sin 2 y+\sin 2 z \\ =-4 \sin x \sin y \sin z \end{array} $

Now, $\frac{\sin 2 x+\sin 2 y+\sin 2 z}{\sin (-x) \sin (-y) \sin (-z)}$

$ \Rightarrow \quad \frac{-4 \sin x \sin y \sin z}{-\sin x \sin y \sin z}=4 $

(D) → (iv)

$ \begin{aligned} & \text { Given, } \cos (3 x+5)+7 \\ & \therefore \quad \text { Period }=\frac{2 \pi}{|3|} \\ & \left\{\because \text { Period of } \cos (a x+b)+c=\frac{2 \pi}{|a|}\right\} \\ & \text { Period }=\frac{2 \pi}{3} \end{aligned} $

Given, $\cos \left(\frac{\alpha-\beta}{2}\right)=2 \cos \left(\frac{\alpha+\beta}{2}\right)$

$ \therefore \quad \frac{\cos \left(\frac{\alpha-\beta}{2}\right)}{\cos \left(\frac{\alpha+\beta}{2}\right)}=\frac{2}{1} $

Now, using componendo and dividendo,

$ \begin{aligned} & \Rightarrow \quad \frac{\cos \left(\frac{\alpha-\beta}{2}\right)+\cos \left(\frac{\alpha+\beta}{2}\right)}{\cos \left(\frac{\alpha-\beta}{2}\right)-\cos \left(\frac{\alpha+\beta}{2}\right)}=\frac{2+1}{2-1} \\ & \Rightarrow \quad \frac{2 \cos \frac{\alpha}{2} \cos \frac{\beta}{2}}{2 \sin \frac{\alpha}{2} \sin \frac{\beta}{2}}=3 \\ & \Rightarrow \quad \tan \frac{\alpha}{2} \tan \frac{\beta}{2}=\frac{1}{3} \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} & \quad \begin{aligned} \cos x-\sin x & =\sqrt{a} \sin x \\ \Rightarrow\,\,\,\,\,\,\,\,\,\,\,\cos x & =\sin x(\sqrt{a}+1) \end{aligned} \\ & \cos x=\frac{\sin x(\sqrt{a}+1)(\sqrt{a}-1)}{(\sqrt{a}-1)} \\ &(\sqrt{a}-1) \cos x=a \sin x-\sin x \\ & \sqrt{a} \cos x=a \sin x+\cos x-\sin x \\ & \therefore \quad a \sin x+\cos x-\sin x=\sqrt{a} \cos x \end{aligned} \end{aligned} $

A. Period of $\sin ^2 x$ is $\pi$

$ \begin{aligned} f(x) & =\sin ^2 x \\ f(\pi+x) & =\sin ^2(\pi+x) \\ & =(-\sin x)^2 \\ & =\sin ^2 x=f(x) \end{aligned} $

B. $\frac{\pi}{3}(\sqrt{3} \cos 3 x+\sin 3 x)$

$ \begin{aligned} \text { Maximum value } & =\frac{\pi}{3} \sqrt{(\sqrt{3})^2+(1)^2} \\ & =\frac{\pi}{3} \times 2=\frac{2 \pi}{3} \end{aligned} $

C. We have, $\sin \frac{x}{3}+\cos \frac{x}{2}$

Period of $\sin x$ is $6 \pi$ and period of $\cos \frac{x}{2}$ is $4 \pi$.

∴ Period of $\sin \frac{x}{3}+\cos \frac{x}{2}$ is LCM of $6 \pi$ and $4 \pi$ i.e., $12 \pi$.

D. We have, $y=|\sin x|$ and $y=1$

$ \begin{array}{lrl} \Rightarrow & 1 & =|\sin x| \\ \Rightarrow & \sin x & = \pm 1 \\ \Rightarrow & & x = \frac{\pi}{2} \text { in }[0, \pi] \end{array} $

∴ Intersection part is $\frac{\pi}{2}$.

$ \begin{aligned} & \text { We have, } \cot \frac{A}{2}=\sqrt{\frac{1+a}{1-a}} \cot \frac{\theta}{2} \\ & \Rightarrow \quad \cot ^2 \frac{A}{2}=\left(\frac{1+a}{1-a}\right) \cot ^2 \frac{\theta}{2} \\ & \Rightarrow \quad\left(\frac{\cos ^2 \frac{A}{2}}{\sin ^2 \frac{A}{2}}\right)\left(\frac{1-a}{1+a}\right)=\frac{\cos ^2 \theta / 2}{\sin ^2 \theta / 2} \\ & \Rightarrow \quad \frac{(1+\cos A)(1-a)}{(1-\cos A)(1+a)}=\frac{1+\cos \theta}{1-\cos \theta} \\ & \Rightarrow \frac{1-a-a \cos A+\cos A}{1+a-a \cos A-\cos A}=\frac{1+\cos \theta}{1-\cos \theta} \end{aligned} $

Apply componendo and dividendo,

$ \begin{array}{rlrl} \Rightarrow & & \frac{2(1-a \cos A)}{2(\cos A-a)} & =\frac{2}{2 \cos \theta} \\ \Rightarrow & \cos \theta & =\frac{\cos A-a}{1-a \cos A} \end{array} $

$ \begin{aligned} &\text { We have, } \sin \theta \cosh \alpha=\tan x\\ &\begin{gathered} \cos \theta \sinh \alpha=\sec x \\ \cos ^2 \theta \sin ^2 \mathrm{~h} \alpha-\sin ^2 \theta \cos ^2 \mathrm{~h} \alpha \\ =\sec ^2 x-\tan ^2 x \\ \cos ^2 \theta \sin ^2 \mathrm{~h} \alpha-\sin ^2 \theta \cos ^2 \mathrm{~h} \alpha=1 \\ \cos ^2 \theta \sin ^2 \mathrm{~h} \alpha-\cos ^2 \mathrm{~h} \alpha \\ +\cos ^2 \theta \cos ^2 \mathrm{~h} \alpha=1 \\ \cos ^2 \theta\left(\sin ^2 \mathrm{~h} \alpha+\cos ^2 \mathrm{~h} \alpha\right)-\cos ^2 \mathrm{~h} \alpha=1 \\ \cos ^2 \theta \cosh 2 \alpha-\cos ^2 \mathrm{~h} \alpha=1 \\ \cos ^2 \theta \cosh 2 \alpha-\left(\frac{\cosh 2 \alpha+1}{2}\right)=1 \\ 2 \cos ^2 \theta \cosh 2 \alpha-\cosh 2 \alpha-1=2 \\ \cosh 2 \alpha\left(2 \cos ^2 \theta-1\right)=3 \\ \Rightarrow \quad \cos 2 \theta \cosh 2 \alpha=3 \end{gathered} \end{aligned} $

$ \begin{aligned} &\text { Maximum value of } 3 \cos \theta-4 \sin \theta\\ &\begin{array}{ll} & a=\sqrt{3^2+(-4)^2}=5 \\ \therefore & \alpha=5 \sin ^2 \theta \cos ^3 \theta \\ & \beta=5 \sin ^3 \theta \cos ^2 \theta \end{array} \end{aligned} $

$ \begin{aligned} &\text { Now, }\\ &\begin{aligned} & \alpha^2+\beta^2=5^2\left(\sin ^4 \theta \cos ^6 \theta+\sin ^6 \theta \cos ^4 \theta\right) \\ & =25\left(\cos ^2 \theta+\sin ^2 \theta\right) \sin ^4 \theta \cos ^4 \theta \\ & =25 \sin ^4 \theta \cos ^4 \theta \\ & \therefore \quad\left(\alpha^2+\beta^2\right)^5=\left(25 \sin ^4 \theta \cos ^4 \theta\right)^5 \\ & \Rightarrow\left(\alpha^2+\beta^2\right)^{5 / 2}=\left(5 \sin ^2 \theta \cos ^2 \theta\right)^5 \\ & \quad \alpha \beta=25 \sin ^5 \theta \cos ^5 \theta \\ & \Rightarrow \quad(\alpha \beta)^2=\left(5^2 \sin ^5 \theta \cos ^5 \theta\right)^2 \\ & \therefore \quad \sqrt{\frac{\left(\alpha^2+\beta^2\right)^5}{(\alpha \beta)^4}}=\frac{\left(5 \sin ^2 \theta \cos ^2 \theta\right)^5}{\left(5^2 \sin ^5 \theta \cos ^5 \theta\right)^2} \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \quad=\frac{5^5 \sin ^{10} \theta \cos ^{10} \theta}{5^4 \sin ^{10} \theta \cos ^{10} \theta}=5 \end{aligned} \end{aligned} $

Given,

$\sin A=\frac{11}{61} A$ lies in 2nd quadrant

$\cos B=\frac{-7}{25} B$ lies in 3rd quadrant

$\cos A=\frac{-60}{61}$ and $\sin B=\frac{-24}{25}$

$ \begin{array}{r} \because \sin (A-B)=\sin A \cos B-\cos A \sin B \\ \quad=\left(\frac{11}{61}\right)\left(\frac{-7}{25}\right)-\left(\frac{-60}{61}\right)\left(\frac{-24}{25}\right) \end{array} $

$ \begin{aligned} & \sin (A-B)<0 \\ & \cos (A-B)=\cos A \cos B+\sin A \sin B \\ & =\left(\frac{-7}{25}\right)\left(\frac{-60}{61}\right)+\left(\frac{11}{61}\right)\left(\frac{-24}{25}\right) \end{aligned} $

$ \cos (A-B)>0 $

$\because A-B$ lie in 4th quadrant

$ \begin{aligned} & \sin (A+B)=\left(\frac{11}{61}\right)\left(\frac{-7}{25}\right)+\left(\frac{-60}{61}\right)\left(\frac{-24}{25}\right) \\ & \sin (A+B)>0 \\ & \cos (A+B)=\left(\frac{-7}{25}\right)\left(\frac{-60}{61}\right)-\left(\frac{11}{61}\right)\left(\frac{-24}{25}\right) \\ & \cos (A+B)>0 \end{aligned} $

$\therefore(A+B)$ lies in Ist quadrant

$ \begin{aligned} & \text { Given, } \\ & \cos \left(\frac{\pi}{4}-x\right) \cos 2 x+\sin x \sin 2 x \sec x \\ & =\cos x \sin 2 x \sec x+\cos \left(\frac{\pi}{4}+x\right) \cos 2 x \\ & \cos 2 x\left[\cos \left(\frac{\pi}{4}-x\right)+\cos \left(\frac{\pi}{4}+x\right)\right] \\ & =\sec x \sin 2 x(\cos x-\sin x) \\ & \cos 2 x\left(2 \sin \frac{\pi}{4} \sin x\right) \\ & =\sec x \sin 2 x(\cos x-\sin x) \\ & \begin{array}{c} \frac{2}{\sqrt{2}}\left(\cos ^2 x-\sin ^2 x\right) \sin x \\ =\sec x(2 \sin x \cos x)(\cos x-\sin x) \\ \cos x+\sin x=\sqrt{2} \end{array} \\ & \begin{array}{r} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\because\,\, sin \,\,x=\cos x=\frac{1}{\sqrt{2}} \\ \sec x=\sqrt{2} \end{array} \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} & \sin ^4 \frac{\pi}{8}+\cos ^4 \frac{3 \pi}{8}-\sin ^4 \frac{3 \pi}{8}+\sin ^4 \frac{5 \pi}{8} \\ & +\cos ^4 \frac{7 \pi}{8}-\sin ^4 \frac{7 \pi}{8} \\ & =\sin ^4 \frac{\pi}{8}+\cos ^4 \frac{3 \pi}{8}-\sin ^4 \frac{3 \pi}{8} \\ & +\sin ^4\left(\pi-\frac{3 \pi}{8}\right)+\cos ^4 \frac{7 \pi}{8}-\sin ^4\left(\pi-\frac{\pi}{8}\right) \\ & =\sin ^4 \frac{\pi}{8}+\cos ^4 \frac{3 \pi}{8}-\sin ^4 \frac{3 \pi}{8}+\sin ^4 \frac{3 \pi}{8} \\ & +\cos ^4 \frac{7 \pi}{8}-\sin ^4 \frac{\pi}{8}=\cos ^4 \frac{3 \pi}{8}+\cos ^4 \frac{7 \pi}{8} \\ & =\cos ^4 \frac{3 \pi}{8}+\cos ^4\left(\pi-\frac{\pi}{8}\right) \\ & =\cos ^4 \frac{3 \pi}{8}+\left(-\cos \frac{\pi}{8}\right)^4 \\ & =\cos ^4 \frac{3 \pi}{8}+\cos ^4 \frac{\pi}{8} \end{aligned} \end{aligned} $

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

$ \begin{aligned} & \text { Reason In } \triangle P Q R, \\ & P+Q+R=180^{\circ} \\ & \Rightarrow \quad \frac{P}{2}+\frac{Q}{2}+\frac{R}{2}=90^{\circ} \\ & \Rightarrow \quad \frac{P}{2}+\frac{Q}{2}=90^{\circ}-\frac{R}{2} \\ & \Rightarrow \quad \tan \left(\frac{P}{2}+\frac{Q}{2}\right)=\tan \left(90-\frac{R}{2}\right) \\ & \Rightarrow \quad \tan \frac{P}{2}+\tan \frac{Q}{2} \\ & \Rightarrow \quad \frac{P}{1-\tan \frac{P}{2} \tan \frac{Q}{2}}=\cot \frac{R}{2} \\ & \Rightarrow\left(\tan \frac{P}{2}+\tan \frac{Q}{2}\right) \tan \frac{R}{2}=1-\tan \frac{P}{2} \tan \frac{Q}{2} \\ & \Rightarrow \tan \frac{P}{2} \tan \frac{Q}{2}+\tan \frac{Q}{2} \tan \frac{R}{2} +\tan \frac{R}{2} \tan \frac{P}{2}=1 \end{aligned} $

So, reason is true.

Assertion

We have,

$ \begin{aligned} & A=15^{\circ}, B=17^{\circ}, C=13^{\circ} \\ \Rightarrow & A+B+C=45^{\circ} \\ \Rightarrow & 2 A+2 B+2 C=90^{\circ} \\ \therefore & \frac{P}{2}=2 A, \frac{Q}{2}=2 B, \frac{R}{2}=2 C \\ \therefore & \tan 2 A \tan 2 B+\tan 2 B \tan 2 C +\tan 2 C \tan 2 A=1 \\ \Rightarrow & \frac{1}{\cot 2 A \cot 2 B}+\frac{1}{\cot 2 B \cot 2 C} +\frac{1}{\cot 2 C \cot 2 A}=1 \\ \Rightarrow & \frac{\cot 2 C+\cot 2 A+\cot 2 B}{\cot 2 A \cot 2 B \cot 2 C}=1 \\ \Rightarrow & \cot 2 A+\cot 2 B+\cot 2 C \\ = & \cot 2 A \cot 2 B \cot 2 C \\ \therefore & \text { Assertion is true. } \end{aligned} $