INVERSE TRIGNOMETRIC FUNCTIONS — Maths STD 12 Science — Question

$[\because\ \tan^{-1}(-\text{x})=-\tan^{-1}\text{x},\ \text{x}\in\text{R}]$

$=-\Big[\cos^{-1}\frac{1-3^2}{1+3^2}\Big]$

$\Big[\because\ 2\tan^{-1}\text{x}=\cos^{-1}\frac{1-\text{x}^2}{1+\text{x}^2},\ \text{x}\geq0\Big]$

$=-\Big[\cos^{-1}\Big(\frac{-8}{10}\Big)\Big]=-\Big[\cos^{-1}\Big(\frac{-4}{5}\Big)\Big]$

$=-\Big[\pi-\cos^{-1}\Big(\frac{4}{5}\Big)\Big]$

$\{\because\ \cos^{-1}(-\text{x})=\pi-\cos^{-1}\text{x},\ \text{x}\in[-1,1]\}$

$=-\pi+\cos^{-1}\Big(\frac{4}{5}\Big)$

$\Big[\text{let}\ \cos^{-1}\Big(\frac{4}{5}\Big)=\theta\Rightarrow\ \cos\theta=\frac{4}{5}\Rightarrow\ \tan\theta=\frac{3}{4}\Rightarrow\ \theta=\tan^{-1}\frac{3}{4}\Big]$

$=-\pi+\tan^{-1}\Big(\frac{3}{4}\Big)=-\pi+\Big[\frac{\pi}{2}-\cot^{-1}\Big(\frac{3}{4}\Big)\Big]$

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

$=-\frac{\pi}{2}+\tan^{-1}\Big(\frac{-4}{3}\Big)$

$[\because\ \tan^{-1}(-\text{x})=-\tan^{-1}\text{x}]$

= RHS (Hence proved)