Q.Prove that $\tan^{-1}x+\cot^{-1}x=\dfrac{\pi}{2}$ for every $x\in\mathbb{R}$.

Apply the complementary identity $\sin^{-1}t + \cos^{-1}t = \dfrac{\pi}{2}$ to both $x$ and $y$; the answer is $\dfrac{\pi}{3}$.

This is a standard CBSE/NCERT Class 12 inverse trigonometric functions identity, tested here in the WBCHSE Semester-III paper.

For every real $t \in [-1,1]$ we have the key identity

$\sin^{-1}t + \cos^{-1}t = \frac{\pi}{2}.$

Write it once for $x$ and once for $y$ and add:

$(\sin^{-1}x + \cos^{-1}x) + (\sin^{-1}y + \cos^{-1}y) = \frac{\pi}{2} + \frac{\pi}{2} = \pi.$

Regroup:

$(\sin^{-1}x + \sin^{-1}y) + (\cos^{-1}x + \cos^{-1}y) = \pi.$

We are given $\sin^{-1}x + \sin^{-1}y = \dfrac{2\pi}{3}$, so

$\cos^{-1}x + \cos^{-1}y = \pi - \frac{2\pi}{3} = \frac{\pi}{3}.$

Use $\cot^{-1}t = \pi/2 - \tan^{-1}t$ to convert the given sum.

For every real $t$, $\tan^{-1}t+\cot^{-1}t=\pi/2$, so $\cot^{-1}t=\pi/2-\tan^{-1}t$.

$\cot^{-1}x+\cot^{-1}y = \left(\frac{\pi}{2}-\tan^{-1}x\right)+\left(\frac{\pi}{2}-\tan^{-1}y\right) = \pi - (\tan^{-1}x+\tan^{-1}y)$

Given $\tan^{-1}x+\tan^{-1}y = \dfrac{4\pi}{5}$, substitute:

$\cot^{-1}x+\cot^{-1}y = \pi - \frac{4\pi}{5} = \frac{\pi}{5}$

Use the identity $\sin^{-1}x + \cos^{-1}x = \dfrac{\pi}{2}$.

For any $x \in [-1,1]$:

$\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$

Here $x = \dfrac{1}{5} \in [-1,1]$, so:

$\sin^{-1}\frac{1}{5} + \cos^{-1}\frac{1}{5} = \frac{\pi}{2}$

Therefore:

$\sin\left(\sin^{-1}\frac{1}{5} + \cos^{-1}\frac{1}{5}\right) = \sin\frac{\pi}{2} = 1$

This is a standard inverse-trig identity, valid for $|x|\ge 1$.

Let $\csc^{-1}x=\theta$, so $x=\csc\theta$, i.e. $\dfrac1x=\sin\theta$, giving $\theta=\sin^{-1}\dfrac1x$.

Similarly $\sec^{-1}x=\cos^{-1}\dfrac1x$.

Using the identity $\sin^{-1}y+\cos^{-1}y=\dfrac{\pi}{2}$ with $y=\dfrac1x$:

$\csc^{-1}x+\sec^{-1}x=\sin^{-1}\dfrac1x+\cos^{-1}\dfrac1x=\dfrac{\pi}{2}$.

Use $\tan(\pi/2-\theta)=\cot\theta$, then $\cot(\tan^{-1}(1/3))$ is the reciprocal of $1/3$.

$\tan\left(\dfrac{\pi}{2}-\tan^{-1}\dfrac13\right)=\cot\left(\tan^{-1}\dfrac13\right)=\dfrac{1}{1/3}=3$