Question

If $\tan \left(\frac{x}{2}\right)=co\sec x-\sin x$, then the value of ${\tan }^2\left(\frac{x}{2}\right)$ is

• A. $2-\sqrt{5}$
• B. $2+\sqrt{5}$
• C. $-2-\sqrt{5}$
• D. $-2+\sqrt{5}$

Answer 1

The correct option is D $-2+\sqrt{5}$

$\sin x=\frac{2\tan x/2}{1+{\tan }^{2}x/2}$
Let $\tan \left(\frac{x}{2}\right)=y$
$\therefore \tan \left(\frac{x}{2}\right)=co\sec x-\sin x=\frac{1}{\sin x}-\sin x$
$\Rightarrow y=\frac{1+y^2}{2y}-\frac{2y}{1+y^2}$
$\Rightarrow 2y^2+2y^4=1+y^4-2y^2=0$
$1+y^4-2y^2-2y^4-2y^2=0$
$\Rightarrow y^2=\frac{-4\pm \sqrt{16+4}}{2}$
$=-2\pm \sqrt{5}$
$y^2=-2+\sqrt{5};-2-\sqrt{5}$
But $y^2\ne -2-\sqrt{5}$
$y^2={\tan }^2\left(x/2\right)=-2+\sqrt{5}$