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Question: Answered & Verified by Expert
$4 \tan ^{-1} \frac{1}{5}-\tan ^{-1} \frac{1}{70}+\tan ^{-1} \frac{1}{99}=$
MathematicsInverse Trigonometric FunctionsJEE Main
Options:
  • A $\frac{\pi}{2}$
  • B $\frac{\pi}{3}$
  • C $\frac{\pi}{4}$
  • D None of these
Solution:
1249 Upvotes Verified Answer
The correct answer is: $\frac{\pi}{4}$
$4 \tan ^{-1} \frac{1}{5}-\tan ^{-1} \frac{1}{70}+\tan ^{-1} \frac{1}{99}$
$\begin{aligned} & =2 \tan ^{-1}\left[\frac{\frac{2}{5}}{1-\frac{1}{25}}\right]-\tan ^{-1} \frac{1}{70}+\tan ^{-1} \frac{1}{99} \\ & =2 \tan ^{-1}\left(\frac{5}{12}\right)-\tan ^{-1} \frac{1}{70}+\tan ^{-1} \frac{1}{99} \\ & =\tan ^{-1}\left[\frac{\frac{5}{6}}{1-\frac{25}{144}}\right]-\tan ^{-1} \frac{1}{70}+\tan ^{-1} \frac{1}{99} \\ & =\tan ^{-1}\left(\frac{120}{119}\right)-\tan ^{-1} \frac{1}{70}+\tan ^{-1} \frac{1}{99} \\ & =\tan ^{-1}\left(\frac{120}{119}\right)+\tan ^{-1}\left[\frac{\frac{1}{99}-\frac{1}{70}}{1+\frac{1}{99} \cdot \frac{1}{70}}\right] \\ & =\tan ^{-1}\left(\frac{120}{119}\right)+\tan ^{-1}\left(\frac{-29}{6931}\right) \\ & =\tan ^{-1} \frac{120}{119}-\tan ^{-1} \frac{29}{6931}=\tan ^{-1} \frac{120}{119}-\tan ^{-1} \frac{1}{239} \\ & =\tan ^{-1}\left[\frac{\frac{120}{119}-\frac{1}{239}}{1+\frac{120}{119} \times \frac{1}{239}}\right]=\tan ^{-1}(1)=\frac{\pi}{4}\end{aligned}$

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