A) \[{{\cos }^{-1}}\sqrt{(\sin \theta )}\]
B) \[{{\sin }^{-1}}\sqrt{(\sin \theta )}\]
C) \[\sqrt{\theta }\]
D) None of these
Correct Answer: B
Solution :
Let \[{{\cos }^{-1}}(\cos \theta +i\sin \theta )=x+iy\] ...(i) \[\Rightarrow \] \[\cos \theta +i\sin \theta =\cos (x+iy)\] \[=\cos x\cos (iy)-\sin x\sin (iy)\] \[\Rightarrow \] \[\cos \theta +i\sin \theta =\cos x\cosh y-i\sin x\sinh y\] On comparing the real and imaginary parts, we get \[\cos \theta =\cos x\cosh y,\sin \theta =-\sin x.\sinh y\] \[\Rightarrow \] \[\cosh y=\frac{\cos \theta }{\cos x}\] ?.(ii) \[\sinh y=-\frac{\sin \theta }{\sin x}\] ?..(iii) On squaring and then subtracting, we get \[{{\cosh }^{2}}y-{{\sinh }^{2}}y=\frac{{{\cos }^{2}}\theta }{{{\cos }^{2}}x}-\frac{{{\sin }^{2}}\theta }{{{\sin }^{2}}x}\] \[\Rightarrow \] \[1=\frac{{{\cos }^{2}}\theta {{\sin }^{2}}x-{{\sin }^{2}}\theta {{\cos }^{2}}x}{{{\cos }^{2}}x{{\sin }^{2}}x}\] \[\Rightarrow \] \[(1-{{\sin }^{2}}x){{\sin }^{2}}x=(1-{{\sin }^{2}}\theta ){{\sin }^{2}}x\] \[-{{\sin }^{2}}\theta (1-{{\sin }^{2}}x)\] \[\Rightarrow \] \[{{\sin }^{2}}x-{{\sin }^{4}}x={{\sin }^{2}}x-{{\sin }^{2}}\theta {{\sin }^{2}}x\] \[-{{\sin }^{2}}\theta +{{\sin }^{2}}\theta {{\sin }^{2}}x\] \[\Rightarrow \] \[{{\sin }^{4}}x={{\sin }^{2}}\theta \Rightarrow {{\sin }^{2}}x=\sin \theta \] \[\Rightarrow \] \[\sin x=\sqrt{\sin \theta }\] \[\Rightarrow \] \[x={{\sin }^{-1}}\sqrt{\sin \theta }\]
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