Question

cot⁻¹ 1+ tan⁻¹ 2 + cot⁻¹ 1/3=π,Prove it.

Answer 1

Proof :

$\mathrm{L.H.S.=cot^{-1}1+tan^{-1}2+cot^{-1}\frac{1}{3}}$

$\mathrm{=\frac{\pi}{4}+tan^{-1}2+\frac{\pi}{2}-tan^{-1}\frac{1}{3}}$

$\mathrm{=\frac{3\pi}{4}+tan^{-1}2-tan^{-1}\frac{1}{3}}$

$\mathrm{=\frac{3\pi}{4}+tan^{-1}\frac{2-\frac{1}{3}}{1+\frac{2}{3}}}$

$\mathrm{=\frac{3\pi}{4}+tan^{-1}1}$

$\mathrm{=\frac{3\pi}{4}+\frac{\pi}{4}}$

$\mathrm{=\frac{4\pi}{4}}$

$\mathrm{=\pi=R.H.S.}$

Hence, proved.

Answer 2

Answer:

$\huge\underline\mathfrak\purple{Solution}$