Q65 / 160 · TS EAMCET 2021 (06 Aug Shift 1) · Mathematics

$2 \tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{7}\right)$ is equal to

  • A
    $\tan ^{-1}\left(\frac{49}{29}\right)$
  • B
    $\frac{\pi}{2}$
  • C
    0
  • Correct answer
    D
    $\frac{\pi}{4}$

Solution

$\begin{aligned} & 2 \tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{7}\right)=\tan ^{-1}\left(\frac{2\left(\frac{1}{3}\right)}{1-\left(\frac{1}{9}\right)}\right) +\tan ^{-1}\left(\frac{1}{7}\right)\\ & =\tan ^{-1}\left(\frac{3}{4}\right)+\tan ^{-1}\left(\frac{1}{7}\right) \\ & =\tan ^{-1}\left(\frac{\left(\frac{3}{4}\right)+\left(\frac{1}{7}\right)}{1-\left(\frac{3}{4}\right)\left(\frac{1}{7}\right)}\right)=\tan ^{-1}\left(\frac{25}{25}\right)=\frac{\pi}{4}\end{aligned}$

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

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