Question

The value of $\log \left(\tan \left(\frac{\mathrm{\pi }}{4}+\frac{ix}{2}\right)\right)$ is

• A. $i{\tan }^{-1}\left(\sin hx\right)$
• B. $i{\tan }^{-1}\left(\cos hx\right)$
• C. None of these

Answer 1

The correct option is A

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

Explanation for the correct answer:

Step 1: Apply trigonometric identities to simplify

Let $S=\log \left(\tan \left(\frac{\mathrm{\pi }}{4}+\frac{ix}{2}\right)\right)$

$\Rightarrow$ $S=\log \left(\frac{1+\tan \left({\displaystyle \frac{ix}{2}}\right)}{1-\tan \left({\displaystyle \frac{ix}{2}}\right)}\right)$ $...\left[\because \tan \left(A+B\right)=\frac{\tan A+\tan B}{1-\tan A\tan B}\right]$

$\Rightarrow$ $S=\log \left(1+\tan \left(\frac{ix}{2}\right)\right)-\log \left(1-\tan \left(\frac{ix}{2}\right)\right)$

$\Rightarrow$ $S=\log \left(1+i\tan h\left(\frac{x}{2}\right)\right)-\log \left(1-i\tan h\left(\frac{x}{2}\right)\right)$ $...\left[\because \tan \left(ix\right)=i\tan hx\right]$

Step 2: Use the formula for logarithm of complex numbers

We know that, $\log \left(a+ib\right)=\frac{1}{2}\log \left(a^2+b^2\right)+i{\tan }^{-1}\left(\frac{b}{a}\right)$

$\Rightarrow$ $S=\frac{1}{2}\log \left(1+\tan h^2\left(\frac{x}{2}\right)\right)+i{\tan }^{-1}\left(\tan h\left(\frac{x}{2}\right)\right)-\frac{1}{2}\log \left(1+\tan h^2\left(\frac{x}{2}\right)\right)+i{\tan }^{-1}\left(\tan h\left(\frac{x}{2}\right)\right)$

$\Rightarrow$ $S=2i{\tan }^{-1}\left(\tan h\left(\frac{x}{2}\right)\right)$

Step 3: Use identities of hyperbolic functions to obtain the required value

$\Rightarrow$ $S=i{\tan }^{-1}\left(\frac{2\tan h\left({\displaystyle \frac{x}{2}}\right)}{\left(1-\tan h^2\left({\displaystyle \frac{x}{2}}\right)\right)}\right)$ $...\left[\because \sin h2x=\frac{2\tan hx}{1-\tan h^{2}x}\right]$

$\Rightarrow$ $S=i{\tan }^{-1}\left(\sin hx\right)$

Hence, the value of $\log \left(\tan \left(\frac{\mathrm{\pi }}{4}+\frac{ix}{2}\right)\right)$ is $i{\tan }^{-1}\left(\sin hx\right)$.

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