Solution:

To Prove: $\tan ^{-1} \frac{1}{3}+\sec ^{-1} \frac{\sqrt{5}}{2}=\frac{\pi}{4}$
Formula Used: $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$ where $x y<1$
Proof:
$\text { LHS }=\tan ^{-1} \frac{1}{3}+\sec ^{-1} \frac{\sqrt{5}}{2} \ldots \text { (1) }$
Let $\sec \theta=\frac{\sqrt{5}}{2}$
Therefore $\theta=\sec ^{-1} \frac{\sqrt{5}}{2} \ldots(2)$
From the figure, $\tan \theta=\frac{1}{2}$
$\Rightarrow \theta=\tan ^{-1} \frac{1}{2} \ldots \text { (3) }$
From (2) and (3),
$\sec ^{-1} \frac{\sqrt{5}}{2}=\tan ^{-1} \frac{1}{2}$
Substituting in (1), we get
$\begin{array}{l}
\text { LHS }=\tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{2} \\
=\tan ^{-1}\left(\frac{\frac{1}{3}+\frac{1}{2}}{1-\left(\frac{1}{3} \times \frac{1}{2}\right)}\right) \\
=\tan ^{-1}\left(\frac{2+3}{6-1}\right) \\
=\tan ^{-1} \frac{5}{5} \\
=\tan ^{-1} 1 \\
=\frac{\pi}{4} \\
=\text { RHS }
\end{array}$
Therefore, LHS = RHS
Hence proved.