A) \[{{\cos }^{-1}}(\sqrt{\sin \theta })\]
B) \[{{\sinh }^{-1}}(\sqrt{\sin \theta })\]
C) \[{{\sin }^{-1}}(\sqrt{\sin \theta })\]
D) \[{{\sin }^{-1}}(\sqrt{\cos \theta })\]
Correct Answer: A
Solution :
\[{{\sin }^{-1}}({{e}^{i\theta }})=x+iy\] Þ \[\sin (x+iy)={{e}^{i\theta }}\] Þ \[\sin x\cosh y+i\cos x\sinh y=\cos \theta +i\sin \theta \] On comparison, \[\cosh y=\frac{\cos \theta }{\sin x}\], \[\sinh y=\frac{\sin \theta }{\cos x}\,\,\,\,\,\,\therefore {{\cosh }^{2}}y-{{\sinh }^{2}}y=1\] Þ \[\frac{1-{{\sin }^{2}}\theta }{1-{{\cos }^{2}}x}-\frac{{{\sin }^{2}}\theta }{{{\cos }^{2}}x}=1\] Þ \[{{\cos }^{2}}x-{{\sin }^{2}}\theta {{\cos }^{2}}x-{{\sin }^{2}}\theta +{{\cos }^{2}}x{{\sin }^{2}}\theta \]= \[{{\cos }^{2}}x-{{\cos }^{4}}x\] Þ \[{{\cos }^{4}}x={{\sin }^{2}}\theta \,\,\,\,\Rightarrow x={{\cos }^{-1}}\sqrt{\sin \theta }\].
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