Q77 / 180 · KCET 2017 · Mathematics
If \( \tan ^{-1} x+\tan ^{-1} y=\frac{4 \pi}{5} \), then \( \cot ^{-1} x+\cot ^{-1} y \) is equal to
- A\( 1 \)
- Correct answerB\( \frac{\Pi}{5} \)
- C\( \frac{2 \pi}{5} \)
- D\( \frac{3 \pi}{5} \)
Solution
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, \( \frac{\Pi}{2}-\cot ^{-1} x+\frac{\Pi}{2}-\cot ^{-1} y=\frac{4 \Pi}{5} \)
\( \Rightarrow \frac{\Pi}{2}+\frac{\Pi}{2}=\frac{4 \Pi}{5}+\cot ^{-1} x+\cot ^{-1} y \)
\( \Rightarrow \cot ^{-1} x+\cot ^{-1} y=\Pi-\frac{4 x}{5}=\frac{\Pi}{5} \)
Hence, the correct answer is option: 'B'.