Question

Find imaginary part of $ {{\sin }^{-1}}(cosec \theta ) $ .

• 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

Answer 1

Answer: (A)

Let $ {{\sin }^{-1}}(cosec \theta )=x+iy $
$ \therefore $ $ cosec \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) $