We have, $\cos \left[\cos ^{-1}\left(\frac{-\sqrt{3}}{2}\right)+\frac{\pi}{6}\right]=\cos \left[\cos ^{-1}\left(\cos \frac{5 \pi}{6}\right)+\frac{\pi}{6}\right] \quad\left[\because \cos \frac{5 \pi}{6}=\frac{-\sqrt{3}}{2}\right]$

$$\begin{aligned} & =\cos \left(\frac{5 \pi}{6}+\frac{\pi}{6}\right) \quad\left\{\because \cos ^{-1} \cos x=x ; x \in[0, \pi]\right\} \\ & =\cos \left(\frac{6 \pi}{6}\right) \\ & =\cos (\pi)=-1 \end{aligned}$$

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

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

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

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

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

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

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