We have, $\sin \left(2 \tan ^{-1} \frac{1}{3}\right)+\cos \left(\tan ^{-1} 2 \sqrt{2}\right)$

$\begin{array}{r}\left.=\sin \left[\sin ^{-1}\left\{\frac{2 \times \frac{1}{3}}{1+\left(\frac{1}{3}\right)^2}\right\}\right]+\cos \left(\cos ^{-1} \frac{1}{3}\right)\right] \\ {\left[\because 2 \tan ^{-1} x=\cos ^{-1} \frac{1}{\sqrt{1+x^2}}\right]} \\ {\left[\because=\sin ^{-1} \frac{2 x}{1+x^2},-1 \leq x \leq 1 \text { and } \tan ^{-1}(2 \sqrt{2})=\cos ^{-1} \frac{1}{3}\right]}\end{array}$

$=\sin \left[\sin ^{-1}\left(\frac{\frac{2}{3}}{1+\frac{1}{9}}\right)\right]+\frac{1}{3} \quad\left\{\because \cos \left(\cos ^{-1} x\right)=x ; x \in[-1,1]\right\}$

$$\begin{aligned} & =\sin \left[\sin ^{-1}\left(\frac{2 \times 9}{3 \times 10}\right)\right]+\frac{1}{3}=\sin \left[\sin ^{-1}\left(\frac{3}{5}\right)\right]+\frac{1}{3} \quad\left[\because \sin \left(\sin ^{-1} x\right)=x\right] \\ & =\frac{3}{5}+\frac{1}{3}=\frac{9+5}{15}=\frac{14}{15} \end{aligned}$$

$$\begin{aligned} &\text { We have, } \quad 2 \tan ^{-1}(\cos \theta)=\tan ^{-1}(2 \operatorname{cosec} \theta)\\ &\begin{aligned} & \Rightarrow \quad \tan ^{-1}\left(\frac{2 \cos \theta}{1-\cos ^2 \theta}\right)=\tan ^{-1}(2 \operatorname{cosec} \theta) \\ & {\left[\because 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)\right]} \\ & \Rightarrow \quad\left(\frac{2 \cos \theta}{\sin ^2 \theta}\right)=(2 \operatorname{cosec} \theta) \\ & \Rightarrow \quad(\cot \theta \cdot 2 \operatorname{cosec} \theta)=(2 \operatorname{cosec} \theta) \Rightarrow \cot \theta=1 \\ & \Rightarrow \quad \cot \theta=\cot \frac{\pi}{4} \Rightarrow \theta=\frac{\pi}{4} \end{aligned} \end{aligned}$$

We have, $\quad \cos \left(2 \tan ^{-1} \frac{1}{7}\right)=\sin \left(4 \tan ^{-1} \frac{1}{3}\right)$

$$\begin{aligned} & \Rightarrow \quad \cos \left[\cos ^{-1}\left(\frac{1-\left(\frac{1}{7}\right)^2}{1+\left(\frac{1}{7}\right)^2}\right)\right]=\sin \left[2 \cdot 2 \tan ^{-1} \frac{1}{3}\right] \quad\left[\because 2 \tan ^{-1} x=\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right] \\ & \Rightarrow \quad \cos \left[\cos ^{-1}\left(\frac{\frac{48}{49}}{\frac{50}{49}}\right)\right]=\sin \left[2 \cdot\left(\tan ^{-1} \frac{\frac{2}{3}}{1-\left(\frac{1}{3}\right)^2}\right)\right] \quad\left[\because 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)\right] \\ & \Rightarrow \quad \cos \left[\cos ^{-1}\left(\frac{48 \times 49}{50 \times 49}\right)\right]=\sin \left[2 \tan ^{-1}\left(\frac{18}{24}\right)\right] \\ & \Rightarrow \quad \cos \left[\cos ^{-1}\left(\frac{24}{25}\right)\right]=\sin \left(2 \tan ^{-1} \frac{3}{4}\right) \\ & \Rightarrow \quad \cos \left[\cos ^{-1}\left(\frac{24}{25}\right)\right]=\sin \left(\sin ^{-1} \frac{2 \times \frac{3}{4}}{1+\frac{9}{16}}\right) \quad\left[\because 2 \tan ^{-1} x=\sin ^{-1} \frac{2 x}{1+x^2}\right] \\ & \Rightarrow \quad \frac{24}{25}=\sin \left(\sin ^{-1} \frac{3 / 2}{25 / 16}\right) \\ & \Rightarrow \quad \frac{24}{25}=\frac{48}{50} \Rightarrow \frac{24}{25}=\frac{24}{25} \\ & \therefore \quad \text { LHS }=\text { RHS } \\ & \text { Hence proved. } \end{aligned}$$

We have, $\quad \cos \left(\tan ^{-1} x\right)=\sin \left(\cot ^{-1} \frac{3}{4}\right)$

$$\begin{array}{lll} \Rightarrow & \cos \left(\cos ^{-1} \frac{1}{\sqrt{x^2+1}}\right)=\sin \left(\sin ^{-1} \frac{4}{5}\right) \\ \text { Let } & \tan ^{-1} x=\theta_1 \quad \Rightarrow \tan \theta_1=\frac{x}{1} \\ \Rightarrow & \cos \theta_1=\frac{1}{\sqrt{x^2+1}} \quad \Rightarrow \theta_1=\cos ^{-1} \frac{1}{\sqrt{x^2+1}} \\ \text { and } & \cot ^{-1} \frac{3}{4}=\theta_2 \quad \Rightarrow \cot \theta_2=\frac{3}{4} \\ \Rightarrow & \sin \theta_2=\frac{4}{5} \quad \Rightarrow \theta_2=\sin ^{-1} \frac{4}{5} \end{array}$$

$$\begin{aligned} &\begin{aligned} & \Rightarrow \quad \frac{1}{\sqrt{x^2+1}}=\frac{4}{5} \\ &\left\{\because \cos \left(\cos ^{-1} x\right)=x, x \in[-1,1] \text { and } \sin \left(\sin ^{-1} x\right)=x, x \in[-1,1]\right\} \end{aligned}\\ &\text { On squaring both sides, we get }\\ &\begin{aligned} 16\left(x^2+1\right) & =25 \\ \Rightarrow\quad 16 x^2 & =9 \\ \Rightarrow\quad x^2 & =\left(\frac{3}{4}\right)^2 \\ \therefore\quad x & = \pm \frac{3}{4}=\frac{-3}{4}, \frac{3}{4} \end{aligned} \end{aligned}$$

$$\begin{aligned} &\text { We have, }\\ &\begin{aligned} \tan ^{-1}\left(\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right) & =\frac{\pi}{4}+\frac{1}{2} \cos ^{-1} x^2 \\ \therefore\quad\text { LHS } & =\tan ^{-1}\left(\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right)\quad\text{..... (i)} \end{aligned} \end{aligned}$$

$$\begin{aligned} {\left[\text { let } x^2\right.} & \left.=\cos 2 \theta=\left(\cos ^2 \theta-\sin ^2 \theta\right)=1-2 \sin ^2 \theta=2 \cos ^2 \theta-1\right] \\ \Rightarrow\quad\cos ^{-1} x^2 & =2 \theta \Rightarrow \theta=\frac{1}{2} \cos ^{-1} x^2 \end{aligned}$$

$$\begin{aligned} & \therefore \quad \begin{aligned} \sqrt{1+x^2} & =\sqrt{1+\cos 2 \theta} \\ & =\sqrt{1+2 \cos ^2 \theta-1}=\sqrt{2} \cos \theta \end{aligned} \\ & \text { and } \quad \begin{aligned} \sqrt{1-x^2} & =\sqrt{1-\cos 2 \theta} \\ & =\sqrt{1-1+2 \sin ^2 \theta}=\sqrt{2} \sin \theta \end{aligned} \end{aligned}$$

$$\begin{aligned} \therefore \quad \mathrm{LHS} & =\tan ^{-1}\left(\frac{\sqrt{2} \cos \theta+\sqrt{2} \sin \theta}{\sqrt{2} \cos \theta-\sqrt{2} \sin \theta}\right) \\ & =\tan ^{-1}\left(\frac{\cos \theta+\sin \theta}{\cos \theta-\sin \theta}\right) \\ & =\tan ^{-1}\left(\frac{1+\tan \theta}{1-\tan \theta}\right)=\tan ^{-1}\left(\frac{\tan \frac{\pi}{4}+\tan \theta}{1-\tan \frac{\pi}{4} \cdot \tan \theta}\right) \\ & =\tan ^{-1}\left[\tan \left(\frac{\pi}{4}+\theta\right)\right] \quad\left[\because \tan (x+y)=\frac{\tan x+\tan y}{1-\tan x \cdot \tan y}\right] \\ & =\frac{\pi}{4}+\theta=\frac{\pi}{4}+\frac{1}{2} \cos ^{-1} x^2 \\ & =\text { RHS } \quad \text { Hence proved. } \end{aligned}$$