We know that, $\tan ^{-1} \tan x=x ; x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ and $\cos ^{-1} \cos x=x ; x \in[0, \pi]$

$$\begin{aligned} & \therefore \quad \tan ^{-1}\left(\tan \frac{5 \pi}{6}\right)+\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right) \\ &= \tan ^{-1}\left[\tan \left(\pi-\frac{\pi}{6}\right)\right]+\cos ^{-1}\left[\cos \left(\pi+\frac{7 \pi}{6}\right)\right] \\ &= \tan ^{-1}\left(-\tan \frac{\pi}{6}\right)+\cos ^{-1}\left(-\cos \frac{7 \pi}{6}\right) \quad[\because \cos (\pi+\theta)=-\cos \theta] \\ &=-\tan ^{-1}\left(\tan \frac{\pi}{6}\right)+\pi-\left[\cos ^{-1} \cos \left(\frac{7 \pi}{6}\right)\right] \\ &\left\{\because \tan ^{-1}(-x)=-\tan ^{-1} x ; x \in R \operatorname{and} \cos ^{-1}(-x)=\pi-\cos ^{-1} x ; x \in[-1,1]\right\} \\ &=-\tan ^{-1}\left(\tan \frac{\pi}{6}\right)+\pi-\cos ^{-1}\left[\cos \left(\pi+\frac{\pi}{6}\right)\right] \quad[\because \cos (\pi+\theta)=-\cos \theta] \\ &=-\tan ^{-1}\left(\tan \frac{\pi}{6}\right)+\pi-\left[\cos ^{-1}\left(-\cos \frac{\pi}{6}\right)\right] \quad\left[\because \cos ^{-1}(-x)=\pi-\cos ^{-1} x\right] \\ &=-\tan ^{-1}\left(\tan \frac{\pi}{6}\right)+\pi-\pi+\cos ^{-1}\left(\cos \frac{\pi}{6}\right) \quad \\ &=-\frac{\pi}{6}+0+\frac{\pi}{6}=0 \quad \end{aligned}$$

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

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