Prove the following

Taking L.H.S = 2 tan-1(-3) = -2 tan-1 3 (∵ tan-1 (-x) = – tan-1 x, x ∈ R)

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

$=-2\left[\frac{\pi}{2}-\tan ^{-1} \frac{1}{3}\right] \quad\left(\because \tan ^{-1} x=\cot ^{-1} \frac{1}{x}, x>0\right)$

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

$=-\pi+\tan ^{-1} \frac{2 \cdot \frac{1}{3}}{1-\left(\frac{1}{3}\right)^{2}} \quad\left(\because 2 \tan ^{-1} x=\tan ^{-1} \frac{2 x}{1-x^{2}}\right)$

$=-\pi+\tan ^{-1} \frac{2 / 3}{8 / 9}=-\pi+\tan ^{-1} \frac{3}{4}$

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

$=-\frac{\pi}{2}-\tan ^{-1} \frac{4}{3} \quad\left(\because \tan ^{-1} x=\cot ^{-1} \frac{1}{x}, x>0\right)$

$=-\frac{\pi}{2}+\tan ^{-1}\left(-\frac{4}{3}\right) \quad\left(\because \tan ^{-1}(-x)=-\tan ^{-1} x, x \in R\right)$

= R.H.S

– Hence Proved.