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 $cosx=0$
$\Rightarrow$ $x=\frac{\pi }{2}$
$\therefore$ 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
${\sin }^{-1}(\cos ec\theta )=\log \left(\cot \frac{\theta }{2}\right)$