Question

If tan^-1 $-\frac{1}{\sqrt{3}}$ + cot^-1 x = $\frac{x}{2}$, then x = __________________.

Answer 1

Disclaimer: The solution is provided for the following question.

If tan⁻¹ $-\frac{1}{\sqrt{3}}$ + cot⁻¹ x = $\frac{\mathrm{\pi }}{2}$, then x = __________________.

Solution:

We know

${\tan }^{-1}x+{\cot }^{-1}x=\frac{\mathrm{\pi }}{2}$, for all x ∈ R

$\therefore {\tan }^{-1}\left(-\frac{1}{\sqrt{3}}\right)+{\cot }^{-1}\left(-\frac{1}{\sqrt{3}}\right)=\frac{\mathrm{\pi }}{2}$ .....(1)

It is given that,

${\tan }^{-1}\left(-\frac{1}{\sqrt{3}}\right)+{\cot }^{-1}x=\frac{\mathrm{\pi }}{2}$ .....(2)

From (1) and (2), we get

$x=-\frac{1}{\sqrt{3}}$

If tan⁻¹ $-\frac{1}{\sqrt{3}}$ + cot⁻¹ x = $\frac{\mathrm{\pi }}{2}$, then x = $\underline{-\frac{1}{\sqrt{3}}}$.