Question

If $\tan \left(\frac{x}{2}\right)=cosecx-\sin x$, then ${\tan }^2\left(\frac{x}{2}\right)$ is equal to

• A. $2-\sqrt{5}$
• B. $\sqrt{5}-2$
• C. $(9-4\sqrt{5})(2+\sqrt{5})$
• D. $(9+4\sqrt{5})(2-\sqrt{5})$

Answer 1

Answer: (B), (C)

The correct options are
B $\sqrt{5}-2$
C $(9-4\sqrt{5})(2+\sqrt{5})$

$\tan \left(\frac{x}{2}\right)=cosecx-\sin x$
The given relation can be written as
$\tan \left(\frac{x}{2}\right)=\frac{1-{\sin }^{2}x}{\sin x}$

$\tan \left(\frac{x}{2}\right)=\frac{{\cos }^{2}x}{\sin x}$

$\Rightarrow 2{\sin }^2\left(\frac{x}{2}\right)=\left[{\cos }^2\right(\frac{x}{2})-{\sin }^2(\frac{x}{2}){]}^2$

$\Rightarrow 2{\tan }^2\left(\frac{x}{2}\right){\sec }^2\left(\frac{x}{2}\right)=[1-{\tan }^2(\frac{x}{2}){]}^2$

$\Rightarrow 2{\tan }^2\left(\frac{x}{2}\right)[1+{\tan }^2(\frac{x}{2}\left)\right]=[1-{\tan }^2(\frac{x}{2}){]}^2$
Put $y={\tan }^2\left(\frac{x}{2}\right)$
$\Rightarrow 2y(1+y)=(1-y{)}^2$
$\Rightarrow y^2+4y-1=0$
$\Rightarrow y=\frac{-4\pm \sqrt{16+4}}{2}=-2+\sqrt{5}$
Since $y>0$, we get
$y=\sqrt{5}-2=\frac{(\sqrt{5}-2{)}^2}{\sqrt{5}-2}\cdot \frac{2+\sqrt{5}}{2+\sqrt{5}}$
$=(9-4\sqrt{5})(2+\sqrt{5})$.