Prove that: $\;$ $2 \cot^{-1} \left(2\right) + \text{cosec}^{-1} \left(\dfrac{5}{3}\right) = \dfrac{\pi}{2}$
$\cot^{-1} \left(x\right) = \tan^{-1} \left(\dfrac{1}{x}\right), \hspace{1cm} x > 0$
$\text{cosec}^{-1} \left(x\right) = \sin^{-1} \left(\dfrac{1}{x}\right), \hspace{1cm} \left|x\right| \geq 1$
$\sin^{-1} \left(x\right) = \tan^{-1} \left(\dfrac{x}{\sqrt{1 - x^2}}\right)$ $\hspace{1cm}$ if $\;$ $0 < x < 1$
$2 \tan^{-1} \left(x\right) = \tan^{-1} \left(\dfrac{2x}{1 - x^2}\right), \;\; x^2 < 1$
$\tan^{-1} \left(x\right) + \tan^{-1} \left(y\right) = \dfrac{\pi}{2}, \;\; \text{if } x \cdot y = 1$
Now, $\;$ $\cot^{-1} \left(2\right) = \tan^{-1} \left(\dfrac{1}{2}\right), \;\;\; \left[2 > 0\right]$
$\text{cosec}^{-1} \left(\dfrac{5}{3}\right) = \sin^{-1} \left(\dfrac{1}{5/3}\right) = \sin^{-1} \left(\dfrac{3}{5}\right), \;\;\; \left|\dfrac{5}{3}\right| \geq 1$
$\begin{aligned}
\therefore \; 2 \cot^{-1} \left(2\right) + \text{cosec}^{-1} \left(\dfrac{5}{3}\right) & = 2 \tan^{-1} \left(\dfrac{1}{2}\right) + \sin^{-1} \left(\dfrac{3}{5}\right) \\\\
& \left[\left(\dfrac{1}{2}\right)^2 < 1, \;\; 0 < \dfrac{3}{5} < 1\right] \\\\
& = \tan^{-1} \left[\dfrac{2 \times \dfrac{1}{2}}{1 - \left(\dfrac{1}{2}\right)^2}\right] + \tan^{-1} \left[\dfrac{3/5}{\sqrt{1 - \left(3/5\right)^2}}\right] \\\\
& = \tan^{-1} \left(\dfrac{1}{3/4}\right) + \tan^{-1} \left(\dfrac{3}{4}\right) \\\\
& = \tan^{-1} \left(\dfrac{4}{3}\right) + \tan^{-1} \left(\dfrac{3}{4}\right) \\\\
& = \dfrac{\pi}{2} \hspace{1cm} \left[\because \dfrac{4}{3} \times \dfrac{3}{4} = 1\right]
\end{aligned}$
Hence proved.