A) \[\log \left( \cot \frac{\theta }{2} \right)\]
B) \[\frac{\pi }{2}\]
C) \[\frac{1}{2}\log \left( \cot \frac{\theta }{2} \right)\]
D) None of these
Correct Answer: A
Solution :
Let \[{{\sin }^{-1}}(\cos ec\theta )=x+iy\] \[\therefore \] \[\cos ec\theta =\sin (x+iy)\] \[=\sin x.\cosh y+i\cos x.\sinh y\] By comparing, we get \[\sin x.\cosh y=\cos ec\theta \] ... (i) and \[\cos x.\cosh y=0\] ...(ii) From Eq. (ii), we get \[cos\text{ }x=0\] \[\Rightarrow \] \[x=\frac{\pi }{2}\] \[\therefore \]From Eq. (i), we get \[\sin \frac{\pi }{2}.\cosh y=\cos ec\theta \] Or \[y={{\cosh }^{-1}}(\cos ec\theta )\] \[\Rightarrow \] \[y=\log (\cos ec\theta +\cot \theta )\] \[=\log \left( \cot \frac{\theta }{2} \right)\] \[\therefore \]Imaginary part of \[{{\sin }^{-1}}(\cos ec\theta )=\log \left( \cot \frac{\theta }{2} \right)\]
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