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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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\( \frac{\Pi}{5} \)
Given that, \( \tan ^{-1} x+\tan ^{-1} y=\frac{4 \pi}{5} \)
We know that, \( \tan ^{-1} x+\cot ^{-1} x=\frac{\Pi}{2} \)
\( \Rightarrow \tan ^{-1} x=\frac{I I}{2}-\cot ^{-1} x \)
So, \( \frac{\Pi}{2}-\cot ^{-1} x+\frac{\Pi}{2}-\cot ^{-1} y=\frac{4 \Pi}{5} \)
\( \Rightarrow \frac{\Pi}{2}+\frac{\Pi}{2}=\frac{4 \Pi}{5}+\cot ^{-1} x+\cot ^{-1} y \)
\( \Rightarrow \cot ^{-1} x+\cot ^{-1} y=\Pi-\frac{4 x}{5}=\frac{\Pi}{5} \)
We know that, \( \tan ^{-1} x+\cot ^{-1} x=\frac{\Pi}{2} \)
\( \Rightarrow \tan ^{-1} x=\frac{I I}{2}-\cot ^{-1} x \)
So, \( \frac{\Pi}{2}-\cot ^{-1} x+\frac{\Pi}{2}-\cot ^{-1} y=\frac{4 \Pi}{5} \)
\( \Rightarrow \frac{\Pi}{2}+\frac{\Pi}{2}=\frac{4 \Pi}{5}+\cot ^{-1} x+\cot ^{-1} y \)
\( \Rightarrow \cot ^{-1} x+\cot ^{-1} y=\Pi-\frac{4 x}{5}=\frac{\Pi}{5} \)
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