If $3 \tan^{-1} x + \cot^{-1} x = \pi$, then $x$ is equal to

$1$

The correct answer is Option (2) → $1$ ##

We have, $3 \tan^{-1} x + \cot^{-1} x = \pi$

Let $3 \tan^{-1} x = \alpha \Rightarrow \tan^{-1} x = \frac{\alpha}{3} \Rightarrow x = \tan \frac{\alpha}{3} \dots \text{(ii)}$

Again, let $\cot^{-1} x = \beta \Rightarrow x = \cot \beta \dots \text{(iii)}$

From Eqs. (ii) and (iii), we get

$\tan \frac{\alpha}{3} = \cot \beta \Rightarrow \tan \frac{\alpha}{3} = \tan \left( \frac{\pi}{2} - \beta \right)$

$\Rightarrow \frac{\alpha}{3} = \frac{\pi}{2} - \beta \Rightarrow \alpha = \frac{3\pi}{2} - 3\beta \dots \text{(iv)}$

From Eq. (i), we get

$3 \tan^{-1} x + \cot^{-1} x = \pi \Rightarrow \alpha + \beta = \pi$

$\Rightarrow \frac{3\pi}{2} - 3\beta + \beta = \pi \quad [\text{from Eq. (iv)}]$

$\Rightarrow 2\beta = \frac{3\pi}{2} - \pi = \frac{\pi}{2} \Rightarrow \beta = \frac{\pi}{4}$

$∴x = \cot \beta = \cot \frac{\pi}{4} = 1$