Q64 / 160 · AP EAMCET 2024 (20 May Shift 1) · Mathematics

$4 \tan ^{-1} \frac{1}{5}-\tan ^{-1} \frac{1}{70}+\tan ^{-1} \frac{1}{99}=$

  • A
    $\frac{\pi}{12}$
  • B
    $\frac{\pi}{6}$
  • Correct answer
    C
    $\frac{\pi}{4}$
  • D
    $\frac{\pi}{3}$

Solution

$4 \tan ^{-1}\left(\frac{1}{5}\right)-\tan ^{-1}\left(\frac{1}{70}\right)+\tan ^{-1}\left(\frac{1}{99}\right)$ $=2\left\{\tan ^{-1}\left(\frac{\frac{2}{5}}{\left.1-\frac{1}{25}\right)}\right)\right\}-\tan ^{-1}\left(\frac{\frac{1}{70}-\frac{1}{99}}{1+\left(\frac{1}{70}\right)\left(\frac{1}{99}\right)}\right)$ $=2 \tan ^{-1} \frac{5}{12}-\tan ^{-1} \frac{29}{6931}=\tan ^{-1}\left(\frac{\frac{5}{6}}{1-\frac{25}{144}}\right)-\tan ^{-1} \frac{1}{239}$ $=\tan ^{-1} \frac{120}{119}-\tan ^{-1} \frac{1}{239}=\tan ^{-1} 1=\frac{\pi}{4}$

Hence, the correct answer is option: 'C'.

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