Q72 / 150 · MHT CET 2022 (05 Aug Shift 1) · Mathematics

$\tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{8}$ has the value

  • Correct answer
    A
    $\frac{\pi}{4}$
  • B
    $\frac{\pi}{6}$
  • C
    $\frac{\pi}{2}$
  • D
    $\frac{\pi}{3}$

Solution

$\begin{aligned} & \tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{8} \\ & =\tan ^{-1}\left(\frac{\frac{1}{3}+\frac{1}{5}}{1-\frac{1}{3} \times \frac{1}{5}}\right)+\tan ^{-1}\left(\frac{\frac{1}{7}+\frac{1}{8}}{1-\frac{1}{7} \times \frac{1}{8}}\right) \\ & =\tan ^{-1}\left(\frac{4}{7}\right)+\tan ^{-1}\left(\frac{3}{11}\right) \\ & =\tan ^{-1}\left(\frac{\frac{4}{7}+\frac{3}{11}}{1-\frac{4}{7} \times \frac{3}{11}}\right)\end{aligned}$ $=\tan ^{-1}\left(\frac{65}{65}\right)=\tan ^{-1}(1)=\frac{\pi}{4}$

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

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