We have, $\tan ^{-1} \sqrt{x(x+1)}+\sin ^{-1} \sqrt{x^2+x+1}=\frac{\pi}{2}\quad\text{.... (i)}$

Let $$\sin ^{-1} \sqrt{x^2+x+1}=\theta$$

$$\begin{array}{ll} \Rightarrow & \sin \theta=\sqrt{\frac{x^2+x+1}{1}} \\ \Rightarrow & \tan \theta=\frac{\sqrt{x^2+x+1}}{\sqrt{-x^2-x}} \quad \left[\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right] \end{array}$$

$$\begin{aligned} \because\quad\theta & =\tan ^{-1} \frac{\sqrt{x^2+x+1}}{\sqrt{-x^2-x}} \\ & =\sin ^{-1} \sqrt{x^2+x+1} \end{aligned}$$

On putting the value of $\theta$ in Eq. (i), we get

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

We know that, $$\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right), x y< 1$$

$$\begin{array}{ll} \therefore & \tan ^{-1}\left[\frac{\sqrt{x(x+1)}+\sqrt{\frac{x^2+x+1}{-x^2-x}}}{1-\sqrt{x(x+1)} \cdot \sqrt{\frac{x^2+x+1}{-x^2-x}}}\right]=\frac{\pi}{2} \\ \Rightarrow & \tan ^{-1}\left[\frac{\sqrt{x^2+x}+\sqrt{\frac{x^2+x+1}{-1\left(x^2+x\right)}}}{1-\sqrt{\left(x^2+x\right) \cdot \frac{\left(x^2+x+1\right)}{-1\left(x^2+x\right)}}}\right]=\frac{\pi}{2} \\ \Rightarrow & \frac{x^2+x+\sqrt{-\left(x^2+x+1\right)}}{\left[1-\sqrt{-\left(x^2+x+1\right.}\right] \sqrt{\left(x^2+x\right)}}=\tan \frac{\pi}{2}=\frac{1}{0} \end{array}$$

$$\begin{aligned} &\begin{array}{lr} \Rightarrow & {\left[1-\sqrt{-\left(x^2+x+1\right)}\right] \sqrt{\left(x^2+x\right)}=0} \\ \Rightarrow & -\left(x^2+x+1\right)=1 \text { or } x^2+x=0 \\ \Rightarrow & -x^2-x-1=1 \text { or } x(x+1)=0 \\ \Rightarrow & x^2+x+2=0 \text { or } x(x+1)=0 \\ \therefore & x=\frac{-1 \pm \sqrt{1-4 \times 2}}{2} \\ \Rightarrow & x=0 \text { or } x=-1 \end{array}\\ &\text { For real solution, we have } x=0,-1 \text {. } \end{aligned}$$

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

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

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