#### show that $2\tan^{-1}\left ( -3 \right )=-\frac{\pi}{2}+\tan ^{-1}\left ( \frac{-4}{3} \right )$

We have to prove ,

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

LHS=$2\tan^{-1}(-3)$ $\left [ Since, \tan^{-1}\left ( -x \right ) = -\tan^{-1}x,x\epsilon R\right ]$

$=-\left [ \cos^{-1}\frac{1-3^{2}}{1+3^{2}} \right ]$ $\left [ Since 2 \tan^{-1}x=\left [\cos^{-1}\frac{1-x^{2}}{1+x^{2}} \right ], x\geq 0\right ]$

$=-\left [ \cos^{-1}\left (\frac{-8}{10} \right ) \right ]$

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

$=-\left [\pi- \cos^{-1}\left (\frac{4}{5} \right ) \right ]$  $=-\left [ since \cos^{-1}(-x)=\pi-\cos^{-1}x,x\epsilon \left [ -1,1 \right ] \right ]$

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

$\left [ let \cos^{-1}\left ( \frac{4}{5} \right ) =0 \Rightarrow \cos \theta = \left ( \frac{4}{5} \right ) \Rightarrow \tan \theta = \left ( \frac{3}{4} \right ) \Rightarrow \theta = \tan^{-1} \left ( \frac{3}{4} \right )\right ]$

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

$=-\frac{\pi}{2}-\cot^{-1}\left ( \frac{3}{4} \right )$

$\left [ Since, \tan^{-1}\left (-x \right )=-\tan^{-1}x \right ]$

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

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

=RHS

Hence Proved.