Question

If the expression
$\frac{[\sin \left(\frac{x}{2}\right)+\cos \left(\frac{x}{2}\right)-i\tan \left(x\right)]}{[1+2i\sin \left(\frac{x}{2}\right)]}$ is real, then set of all possible values of $x$ is

• A. $n\pi +\alpha$
• B. $2n\pi$
• C. $\frac{n\pi }{2}+\alpha$
• D. None of these

Answer 1

The correct option is B $2n\pi$

Since, $\frac{(\sin \frac{x}{2}+\cos \frac{x}{2}-i\tan \frac{x}{2})(1-2i\sin \frac{x}{2})}{1+4{\sin }^2\frac{x}{2}}$
It will be real, if imaginary part is zero
Therefore, $-2\sin \frac{x}{2}(\sin \frac{x}{2}+\cos \frac{x}{2})-\tan x=0$
$\Rightarrow 2\sin \frac{x}{2}(\sin \frac{x}{2}+\cos \frac{x}{2})+\tan x=0$
$\Rightarrow 2\sin \frac{x}{2}\{(\sin \frac{x}{2}+\cos \frac{x}{2})\cos x+\cos \frac{x}{2}\}=0$
Thus $\sin \frac{x}{2}=0$
$\Rightarrow x=2n\pi$