$$ \begin{array}{lr} \text { We have to prove, } & \cot \left(\frac{\pi}{4}-2 \cot ^{-1} 3\right)=7 \\ \Rightarrow & \left(\frac{\pi}{4}-2 \cot ^{-1} 3\right)=\cot ^{-1} 7 \\ \Rightarrow & \left(2 \cot ^{-1} 3\right)=\frac{\pi}{4}-\cot ^{-1} 7 \\ \Rightarrow & 2 \tan ^{-1} \frac{1}{3}=\frac{\pi}{4}-\tan ^{-1} \frac{1}{7} \\ \Rightarrow & 2 \tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{7}=\frac{\pi}{4} \\ \Rightarrow & \tan ^{-1} \frac{2 / 3}{1-(1 / 3)^2}+\tan ^{-1} \frac{1}{7}=\frac{\pi}{4} \\ \Rightarrow & \tan ^{-1} \frac{2 / 3}{8 / 9}+\tan ^{-1} \frac{1}{7}=\frac{\pi}{4} \\ \Rightarrow & \tan ^{-1} \frac{3}{4}+\tan ^{-1} \frac{1}{7}=\frac{\pi}{4} \\ \Rightarrow & \frac{3}{4}+\frac{1}{7} \\ \Rightarrow & \tan ^{-1} \frac{3}{1-\frac{3}{4} \cdot \frac{1}{7}}=\frac{\pi}{4} \\ \Rightarrow & \tan ^{-1} \frac{(21+4) / 28}{(28-3) / 28}=\frac{\pi}{4} \\ \Rightarrow & \tan ^{-1} \frac{25}{25}=\frac{\pi}{4} \\ \Rightarrow & 1=\tan \frac{\pi}{4} \\ \Rightarrow & 1=1 \\ \Rightarrow & \text{LHS = RHS}\quad\text{Hence proved.} \end{array}$$

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

$$\begin{aligned} & =\tan ^{-1}\left(\tan \frac{5 \pi}{6}\right)+\cot ^{-1}\left(\cot \frac{\pi}{3}\right)+\tan ^{-1}(-1) \\ & =\tan ^{-1}\left[\tan \left(\pi-\frac{\pi}{6}\right)\right]+\cot ^{-1}\left[\cot \left(\frac{\pi}{3}\right)\right]+\tan ^{-1}\left[\tan \left(\pi-\frac{\pi}{4}\right)\right] \\ & =\tan ^{-1}\left(-\tan \frac{\pi}{6}\right)+\cot ^{-1}\left(\cot \frac{\pi}{3}\right)+\tan ^{-1}\left(-\tan \frac{\pi}{4}\right) \end{aligned}$$

$\left[\begin{array}{lrl}\because & \tan ^{-1}(\tan x) & =x, x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \\ & \cot ^{-1}(\cot x) & =x, x \in(0, \pi) \\ \text { and } \quad \tan ^{-1}(-x) & =-\tan ^{-1} x\end{array}\right]$

$$\begin{aligned} & =-\frac{\pi}{6}+\frac{\pi}{3}-\frac{\pi}{4}=\frac{-2 \pi+4 \pi-3 \pi}{12} \\ & =\frac{-5 \pi+4 \pi}{12}=-\frac{\pi}{12} \end{aligned}$$

We have, $\tan ^{-1}\left(\tan \frac{2 \pi}{3}\right)=\tan ^{-1} \tan \left(\pi-\frac{\pi}{3}\right)$

$$\begin{array}{lr} =\tan ^{-1}\left(-\tan \frac{\pi}{3}\right) & {\left[\because \tan ^{-1}(-x)=-\tan ^{-1} x\right]} \\ =-\tan ^{-1} \tan \frac{\pi}{3}=-\frac{\pi}{3} & {\left[\because \tan ^{-1}(\tan x)=x, x \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)\right]} \end{array}$$

$\mathrm{LHS}=2 \tan ^{-1}(-3)=-2 \tan ^{-1} 3 \quad\left[\because \tan ^{-1}(-x)=-\tan ^{-1} x, x \in R\right]$

$=-\left[\cos ^{-1} \frac{1-3^2}{1+3^2}\right] \quad\left[\because 2 \tan ^{-1} x=\cos ^{-1} \frac{1-x^2}{1+x^2}, x \geq 0\right]$

$$\begin{aligned} & =-\left[\cos ^{-1}\left(\frac{-8}{10}\right)\right]=-\left[\cos ^{-1}\left(\frac{-4}{5}\right)\right] \\ & =-\left[\pi-\cos ^{-1}\left(\frac{4}{5}\right)\right] \\ & =-\pi+\cos ^{-1}\left(\frac{4}{5}\right)\left[\text { let } \cos ^{-1}\left(\frac{4}{5}\right)=\theta \Rightarrow \cos \theta=\frac{4}{5} \Rightarrow \tan \theta=\frac{3}{4} \Rightarrow \theta=\tan ^{-1} \frac{3}{4}\right] \end{aligned}$$

$$\begin{aligned} & =-\pi+\tan ^{-1}\left(\frac{3}{4}\right)=-\pi+\left[\frac{\pi}{2}-\cot ^{-1}\left(\frac{3}{4}\right)\right] \\ & =-\frac{\pi}{2}-\cot ^{-1} \frac{3}{4}=-\frac{\pi}{2}-\tan ^{-1} \frac{4}{3} \\ & =-\frac{\pi}{2}+\tan ^{-1}\left(\frac{-4}{3}\right) \quad \left[\because \tan ^{-1}(-x)=-\tan ^{-1} x\right]\\ & =\text { RHS }\quad\text{Hence proved.} \end{aligned}$$

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

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

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