$\cos \left[\sin ^{-1}\left(\frac{3}{5}\right)+\cos ^{-1}\left(\frac{12}{13}\right)\right]=$

$\cos \left[\sin ^{-1}\left(\frac{3}{5}\right)+\cos ^{-1}\left(\frac{12}{13}\right)\right]=$
  1. $\frac{36}{65}$
  2. $\frac{12}{65}$
  3. $\frac{33}{65}$
  4. $\frac{3}{65}$

Solution

$\begin{aligned} & \cos \left[\sin ^{-1}\left(\frac{3}{5}\right)+\cos ^{-1}\left(\frac{12}{13}\right)\right] \\ & =\cos \left(\sin ^{-1}\left(\frac{3}{5}\right)\right) \cos \left(\cos ^{-1}\left(\frac{12}{13}\right)\right) \\ & -\sin \left(\sin ^{-1}\left(\frac{3}{5}\right)\right) \sin \left(\cos ^{-1}\left(\frac{12}{13}\right)\right) \\ & =\sqrt{\frac{25-9}{25}} \times\left(\frac{12}{13}\right)-\left(\frac{3}{5}\right) \times \sqrt{\frac{169-144}{169}} \\ & =\left(\frac{4}{5}\right) \times\left(\frac{12}{13}\right)-\left(\frac{3}{5}\right) \times\left(\frac{5}{13}\right) \\ & =\frac{33}{65}\end{aligned}$

Asked in: MHT CET 2024 (11 May Shift 1)

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