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

Find imaginary part of sin⁡−1(cosecθ)\sin^{- 1}(\text{cosec}\theta)

A

log⁡(cot⁡θ2)\log\left( \cot\frac{\theta}{2} \right)

B

π2\frac{\pi}{2}

C

12log⁡(cot⁡θ2)\frac{1}{2}\log\left( \cot\frac{\theta}{2} \right)

D

None

Answer

log⁡(cot⁡θ2)\log\left( \cot\frac{\theta}{2} \right)

Explanation

Solution

Let sin⁡−1(cosecθ)=x+iy\sin^{- 1}(\text{cosec}\theta) = x + iy

∴\therefore cosecθ=sin⁡(x+iy)\text{cosec}\theta = \sin(x + iy) = sin⁡xcos⁡hy+icos⁡xsin⁡hy\sin x{\cos h}y + i\cos x{\sin h}y

By comparing we get, sin⁡xcosh⁡y=cosecθ\sin x\cosh y = \text{cosec}\theta ......(i) and cos⁡xsin⁡hy=0\cos x{\sin h}y = 0 .........(ii)

From (ii), cos⁡x=0\cos x = 0 ⇒ x=π2x = \frac{\pi}{2}

∴\therefore from (i) sin⁡π2.cosh⁡y=cosecθ\sin\frac{\pi}{2}.\cosh y = \text{cosec}\theta or y=cosh⁡−1(cosecθ)y = \cosh^{- 1}(\text{cosec}\theta) =log⁡[cosecθ]\log\lbrack\text{cosec}\theta\rbrack

⇒y\Rightarrow y = log⁡[cosecθ+cot⁡θ]\log\lbrack\text{cosec}\theta + \cot\theta\rbrack = log⁡(cot⁡θ2)\log\left( \cot\frac{\theta}{2} \right)

∴sin⁡−1(cosecθ)\therefore\sin^{- 1}(\text{cosec}\theta) =π2+ilog⁡(cot⁡θ2)\frac{\pi}{2} + i\log\left( \cot\frac{\theta}{2} \right)

Real part = π2,\frac{\pi}{2}, Imaginary part = log⁡(cot⁡θ2)\log\left( \cot\frac{\theta}{2} \right)