A) 3/5
B) 4/5
C) 1
D) 0
Correct Answer: A
Solution :
[a] Let \[{{\sin }^{-1}}(1)+si{{n}^{-1}}\left( \frac{4}{5} \right)={{\sin }^{-1}}x\] Let \[{{\sin }^{-1}}(1)=\theta \Rightarrow sin\theta =1\Rightarrow cos\theta =0\] And \[{{\sin }^{-1}}\left( \frac{4}{5} \right)=\phi \Rightarrow \sin \phi =\left( \frac{4}{5} \right)\] \[\Rightarrow \cos \phi =\sqrt{1-\frac{16}{25}}=\sqrt{\frac{9}{25}}=\frac{3}{5}\] \[\therefore {{\sin }^{-1}}x=\theta +\phi \] \[\Rightarrow x=\sin (\theta +\phi )=sin\theta cos\phi +cos\theta sin\phi \] \[=1\times \frac{3}{5}+0\times \frac{4}{5}\] \[\Rightarrow x=\frac{3}{5}\]
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