Show that 2 ^ -1 (-3)=- 2 + ^ -1 (-4 3 ). - Vidyadip

Question

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

✓

Answer

LHS $=2\tan^{-1}(-3)=-2\tan^{-1}3$$[\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)

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