Question

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

• A. $x=n\pi +2\alpha ,\alpha ={\tan }^{-1}k$ where $k\in (1,2)$
• B. $x=2n\pi +\alpha ,\alpha ={\tan }^{-1}k$ where $k\in (1,2)$
• C. $x=2n\pi +2\alpha ,\alpha ={\tan }^{-1}k$ where $k\in (1,2)$
• D. None of these

Answer 1

The correct option is C $x=2n\pi +2\alpha ,\alpha ={\tan }^{-1}k$ where $k\in (1,2)$

$\frac{(\sin \frac{x}{2}+\cos \frac{x}{2})-i\tan x}{1+2i\sin \frac{x}{2}}$
$=\frac{(\sin \frac{x}{2}+\cos \frac{x}{2}-i\tan x)(1-2i\sin \frac{x}{2})}{1+4{\sin }^2\frac{x}{2}}$
Since, it is real, so imaginary part will be zero.
$\therefore -2\sin \frac{x}{2}\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)-\tan x=0$
$\Rightarrow 2\sin \frac{x}{2}(\sin \frac{x}{2}+\cos \frac{x}{2})\cos x+2\sin \frac{x}{2}\cos \frac{x}{2}=0$
$\Rightarrow \sin \frac{x}{2}[\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)({\cos }^2\frac{x}{2}-{\sin }^2\frac{x}{2})+\cos \frac{x}{2}]=0$
$\therefore \sin \frac{x}{2}=0\Rightarrow x=2n\pi$ ...(1)
Or $\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)\left({\cos }^2\frac{x}{2}-{\sin }^2\frac{x}{2}\right)+\cos \frac{x}{2}=0$
On dividing by ${\cos }^{3}x$, we have
$\left(\tan \frac{x}{2}+1\right)(1-{\tan }^2\frac{x}{2})+(1+{\tan }^2\frac{x}{2})=0$
$\Rightarrow {\tan }^3\frac{x}{2}-\tan \frac{x}{2}-2=0$
Put $\tan \frac{x}{2}=t$ and $f\left(t\right)=t^3-t-2$
Thus $f\left(1\right)=-2<0$ and $f\left(2\right)=4>0$
Thus $f\left(t\right)$ changes sign from negative to positive in the interval $(1,2)$
Therefore let $t=k$ be the root for which $f\left(k\right)=0$ and $kϵ(1,2)$
$\therefore t=k$ or $\tan \frac{x}{2}=k=\tan 2$
$\Rightarrow \frac{x}{2}=n\pi +\alpha$
$\Rightarrow x=2n\pi +2\alpha ,\alpha ={\tan }^{-1}k$ where $k\in (1,2)$