If $\sin ^{-1}\left(\frac{3}{5}\right)+\cos ^{-1}\left(\frac{12}{13}\right)=\sin ^{-1} \alpha$, then $\alpha=$

If $\sin ^{-1}\left(\frac{3}{5}\right)+\cos ^{-1}\left(\frac{12}{13}\right)=\sin ^{-1} \alpha$, then $\alpha=$
  1. $\frac{56}{65}$
  2. $\frac{61}{65}$
  3. $\frac{63}{65}$
  4. $\frac{62}{65}$

Solution

We have $\sin ^{-1}\left(\frac{3}{5}\right)+\cos ^{-1}\left(\frac{12}{13}\right)=\sin ^{-1} \alpha$ $\begin{aligned} & \therefore \tan ^{-1}\left(\frac{3}{4}\right)+\tan ^{-1}\left(\frac{5}{12}\right)=\tan ^{-1}\left(\frac{\alpha}{\sqrt{1-\alpha^2}}\right) \\ & \therefore \tan ^{-1}\left[\frac{\left(\frac{3}{4}\right)+\left(\frac{5}{12}\right)}{1-\left(\frac{3}{4}\right)\left(\frac{5}{12}\right)}\right]=\tan ^{-1}\left(\frac{\alpha}{\sqrt{1-\alpha^2}}\right) \\ & \therefore \tan ^{-1}\left[\frac{\left(\frac{14}{12}\right)}{\left(\frac{11}{16}\right)}\right]=\tan ^{-1}\left(\frac{\alpha}{\sqrt{1-\alpha^2}}\right) \\ & \therefore \frac{56}{33}=\frac{\alpha}{\sqrt{1-\alpha^2}} \Rightarrow(56)^2(1-\alpha)^2=(33)^2 \alpha^2 \\ & \alpha^2=\frac{(56)^2}{(56)^2+(33)^2}=\frac{(56)^2}{(65)^2} \Rightarrow \alpha=\frac{56}{65} \end{aligned}$

Asked in: MHT CET 2021 (21 Sep Shift 2)

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