If tan ^{-1} x+tan ^{-1} y=frac{4 pi}{5},then | vedclass

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(A) We are given that $	an ^{-1} x+	an ^{-1} y=\frac{4 \pi}{5}$.We know the identity $	an ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}$ for any real $x$.This

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$ \frac{\pi}{5} $

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(A) We are given that $	an ^{-1} x+	an ^{-1} y=\frac{4 \pi}{5}$.We know the identity $	an ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}$ for any real $x$.This implies $	an ^{-1} x=\frac{\pi}{2}-\cot ^{-1} x$ and $	an ^{-1} y=\frac{\pi}{2}-\cot ^{-1} y$.Substituting these into the given equation:$(\frac{\pi}{2}-\cot ^{-1} x) + (\frac{\pi}{2}-\cot ^{-1} y) = \frac{4 \pi}{5}$.$\pi - (\cot ^{-1} x + \cot ^{-1} y) = \frac{4 \pi}{5}$.$\cot ^{-1} x + \cot ^{-1} y = \pi - \frac{4 \pi}{5}$.$\cot ^{-1} x + \cot ^{-1} y = \frac{\pi}{5}$.

If $\tan ^{-1} x+\tan ^{-1} y=\frac{4 \pi}{5}$,then $\cot ^{-1} x+\cot ^{-1} y$ is equal to

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