Question

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

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

Answer 1

The correct option is B

$2n\mathrm{\pi }$

Explanation for the correct option:

Step 1: Given data.

An expression $\frac{\sin \left({\displaystyle \frac{x}{2}}\right)+\cos \left({\displaystyle \frac{x}{2}}\right)-i\tan \left(x\right)}{1+2i\sin \left({\displaystyle \frac{x}{2}}\right)}$is given.

Since it is given that the given expression is real. So, the imaginary terms should be equal to zero.

Step 2: Set imaginary terms equal to zero.

$$\begin{gathered} z=\frac{\sin \left({\displaystyle \frac{x}{2}}\right)+\cos \left({\displaystyle \frac{x}{2}}\right)-i\tan \left(x\right)}{1+2i\sin \left({\displaystyle \frac{x}{2}}\right)} \\ =\frac{\sin \left({\displaystyle \frac{x}{2}}\right)+\cos \left({\displaystyle \frac{x}{2}}\right)-i\tan \left(x\right)}{1+2i\sin \left({\displaystyle \frac{x}{2}}\right)}\times \frac{1-2i\sin \left(\frac{x}{2}\right)}{1-2i\sin \left(\frac{x}{2}\right)} \\ =\frac{\left(\sin \left(\frac{x}{2}\right)+\cos \left(\frac{x}{2}\right)-i\tan \left(x\right)\right)\left(1-2i\sin \left(\frac{x}{2}\right)\right)}{1+4{\sin }^2\left(\frac{x}{2}\right)} \end{gathered}$$

Now, imaginary part of $z$ is equal to zero.

$$\begin{gathered} Im\left(z\right)=-2\sin \left(\frac{x}{2}\right)\left(\sin \left(\frac{x}{2}\right)+\cos \left(\frac{x}{2}\right)\right)-\tan x=0 \\ \Rightarrow 2\sin \left(\frac{x}{2}\right)\left(\sin \left(\frac{x}{2}\right)+\cos \left(\frac{x}{2}\right)\right)=\frac{-2\sin \left(\frac{x}{2}\right)\cos \left(\frac{x}{2}\right)}{\cos x} \\ \Rightarrow \sin \left(\frac{x}{2}\right)=0 \\ \Rightarrow x=2n\mathrm{\pi } \end{gathered}$$

So, the intersection of the values of $x$ is $x=2n\mathrm{\pi }$.

Therefore, the value of $x$ is $2n\mathrm{\pi }$.

Hence, option (B) is the correct answer.