Show that `2tan^-1 (-3)=-pi/2+tan^-1(-4/3)`

Explanation

L.H.S.
`=2 \tan ^{-1}(-3)=-2 \tan ^{-1}(3)`
`=-2[\frac{\pi}{2}-\cot ^{-1} 3] \quad(\∵ \tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2})`
`=-2[\frac{\pi}{2}-\tan ^{-1} \frac{1}{3}] \quad( \∵ \tan ^{-1} x=\cot ^{-1} \frac{1}{x}, x>0)`
`=-\pi+2 \tan ^{-1} \frac{1}{3}`
`=-\pi+\tan ^{-1} \frac{2 \cdot \frac{1}{3}}{1-(\frac{1}{3})^{2}}( \∵ 2 \tan ^{-1} x=\tan ^{-1} \frac{2 x}{1-x^{2}})`
`=-\pi+\tan ^{-1} \frac{2 / 3}{8 / 9}=-\pi+\tan ^{-1} \frac{3}{4}`
`=-\pi+\frac{\pi}{2}-\cot ^{-1} \frac{3}{4} \quad(\∵ \tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2})`
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