Q88 / 150 · MHT CET 2021 (23 Sep Shift 2) · Mathematics

$\tan \left(\cos ^{-1}\left(\frac{4}{5}\right)+\tan ^{-1}\left(\frac{2}{3}\right)\right)=$

  • Correct answer
    A
    $\frac{17}{6}$
  • B
    $\frac{17}{3}$
  • C
    $\frac{18}{5}$
  • D
    $\frac{7}{15}$

Solution

$\begin{aligned} & \tan \left[\cos ^{-1}\left(\frac{4}{5}\right)+\tan ^{-1}\left(\frac{2}{3}\right)\right] \\ & =\tan \left[\tan ^{-1}\left(\frac{3}{4}\right)+\tan ^{-1}\left(\frac{2}{3}\right)\right]=\tan \left[\tan ^{-1}\left(\frac{\left(\frac{3}{4}\right)+\left(\frac{2}{3}\right)}{1-\left(\frac{3}{4}\right)+\left(\frac{2}{3}\right)}\right)\right] \\ & =\tan \left[\tan ^{-1}\left(\frac{17}{6}\right)\right]=\frac{17}{6}\end{aligned}$

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

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