Question

The imaginary part of ${\tan }^{-1}\left(\cos \theta +i\sin \theta \right)$

• A. $\tan h^{-1}\left(\sin \theta \right)$
• B. $\tan h^{-1}\left(\infty \right)$
• C. $\frac{1}{2}\tan h^{-1}\left(\sin \theta \right)$
• D. None of these

Answer 1

The correct option is C

$\frac{1}{2}\tan h^{-1}\left(\sin \theta \right)$

Find the imaginary part of the given expression:

Let ${\tan }^{-1}\left(\cos \theta +i\sin \theta \right)=x+iy$ $...\left(i\right)$

$\therefore$ ${\tan }^{-1}\left(\cos \theta -isin\theta \right)=x-iy$ $...\left(ii\right)$

Subtracting $\left(ii\right)$ from $\left(i\right)$

$2iy={\tan }^{-1}(\cos \theta +i\sin \theta )-{\tan }^{-1}\left(\cos \theta -i\sin \theta \right)$

$={\tan }^{-1}\frac{\left(\cos \theta +isin\theta -\cos \theta +i\sin \theta \right)}{\left(1+\left(\cos \theta +i\sin \theta \right)\left(\cos \theta -i\sin \theta \right)\right)}$

$={\tan }^{-1}\left(\frac{2i\sin \theta }{1+{\cos }^2\theta +{\sin }^2\theta }\right)$

$={\tan }^{-1}\left(i\sin \theta \right)$

$=\tan h^{-1}\left(\sin \theta \right)$

$\therefore$ Imaginary part is $\frac{1}{2}\tan h^{-1}\left(\sin \theta \right)$

Hence, option $\left(C\right)$ is the correct answer.