$\cos ^{-1}\left(\frac{4}{5}\right)+\cos ^{-1}\left(\frac{12}{13}\right)=$

$\cos ^{-1}\left(\frac{4}{5}\right)+\cos ^{-1}\left(\frac{12}{13}\right)=$
  1. $\cos ^{-1}\left(\frac{24}{25}\right)$
  2. $\cos ^{-1}\left(\frac{33}{65}\right)$
  3. $\cos ^{-1}\left(\frac{5}{13}\right)$
  4. $\cos ^{-1}\left(\frac{3}{5}\right)$

Solution

$\begin{aligned} & \cos ^{-1}\left(\frac{4}{5}\right)+\cos ^{-1}\left(\frac{12}{13}\right) \\ & =\tan ^{-1}\left(\frac{3}{4}\right)+\tan ^{-1}\left(\frac{5}{12}\right) \\ & =\tan ^{-1}\left[\frac{\left(\frac{3}{4}\right)+\left(\frac{5}{12}\right)}{1-\left(\frac{3}{12}\right)\left(\frac{5}{12}\right)}\right] \\ & =\tan ^{-1}\left(\frac{36+20}{48-15}\right)=\tan ^{-1}\left(\frac{56}{13}\right)=\cos ^{-1}\left(\frac{33}{65}\right) \end{aligned}$

Asked in: MHT CET 2021 (20 Sep Shift 2)

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