Question

Find imaginary part of ${\sin }^{-1}(\cos ec\theta )$

• A. $\log \left(cot\frac{\theta }{2}\right)$
• B. $\frac{\mathrm{\pi }}{2}$
• C. $\frac{1}{2}\log \left(cot\frac{\theta }{2}\right)$
• D. None of these

Answer 1

The correct option is A

$\log \left(cot\frac{\theta }{2}\right)$

Step 1. According to the question

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

$\Rightarrow$ $\cos ec\theta =\sin (x+iy)$

$=\sin x.\cos hy+i\cos x.\sin hy$

Step 2. compare the real and imaginary part,

$\sin x.\cos hy=\cos ec\theta$ ….... (i)

and $\cos x.\cos hy=0$ …....(ii)

Step 3. From Equation (ii), we get

$\cos x=0$
⇒ $x=\frac{\pi }{2}$

Step 4. Put the value of $x$ in equation (i),

$\sin \frac{\pi }{2}.\cos hy=\cos ec\theta$

$\Rightarrow$ $y=\cos h^{-1}(\cos ec\theta )$

$=\log (\cos ec\theta +cot\theta )$ $\left[\mathbf{\because }\mathit{c}\mathit{o}\mathit{s}{\mathit{h}}^{\mathbf{-}\mathbf{1}}\left(cosec\theta \right)\mathbf{}\mathbf{=}\mathbf{}\mathit{l}\mathit{o}\mathit{g}\left(cosec\theta +cot\theta \right)\right]$
$=\log \left(cot\frac{\theta }{2}\right)$ $\left[\mathbf{\because }\mathit{c}\mathit{o}\mathit{s}\mathit{e}\mathit{c}\mathbf{}\mathit{\theta }\mathbf{+}\mathit{c}\mathit{o}\mathit{t}\mathbf{}\mathit{\theta }\mathbf{}\mathbf{=}\mathbf{}\mathit{c}\mathit{o}\mathit{t}\mathbf{}\frac{\mathbf{\theta }}{\mathbf{2}}\right]$

$\therefore$Imaginary part of $si{n}^{-1}(cosec\theta )=\mathit{l}\mathit{o}\mathit{g}\mathbf{}\left(cot\frac{\theta }{2}\right)$

Hence, Option ‘A’ is Correct.