Question

If $\tan \left(\frac{x}{2}\right)=\mathrm{cosec}\left(x\right)-\sin \left(x\right)$, 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

$-2+\sqrt{5}$

The explanation for the correct option

The given trigonometric expression: $\tan \left(\frac{x}{2}\right)=\mathrm{cosec}\left(x\right)-\sin \left(x\right)$.

It is known that, $\sin \left(\theta \right)=\frac{2\tan \left({\displaystyle \frac{\theta }{2}}\right)}{1+{\tan }^2\left(\frac{\theta }{2}\right)}$.

Thus, $\tan \left(\frac{x}{2}\right)=\mathrm{cosec}\left(x\right)-\sin \left(x\right)$

$$\begin{gathered} \Rightarrow \tan \left(\frac{x}{2}\right)=\frac{1}{\sin \left(x\right)}-\sin \left(x\right)\left[\because \mathrm{cosec}\left(\theta \right)=\frac{1}{\sin \left(\theta \right)}\right] \\ \Rightarrow \tan \left(\frac{x}{2}\right)=\frac{1+{\tan }^2\left(\frac{x}{2}\right)}{2\tan \left(\frac{x}{2}\right)}-\frac{2\tan \left(\frac{x}{2}\right)}{1+{\tan }^2\left(\frac{x}{2}\right)} \end{gathered}$$

Let us assume that $m=\tan \left(\frac{x}{2}\right)$.

$$\begin{gathered} \Rightarrow m=\frac{1+m^2}{2m}-\frac{2m}{1+m^2} \\ \Rightarrow m=\frac{{\left(1+m^2\right)}^2-4m^2}{2m\left(1+m^2\right)} \\ \Rightarrow m=\frac{1+2m^2+m^4-4m^2}{2m+2m^3} \\ \Rightarrow 2m^2+2m^4=1-2m^2+m^4 \\ \Rightarrow m^4+4m^2-1=0 \end{gathered}$$

Solve the equation using the quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$.

Thus, $m^2=\frac{-4\pm \sqrt{{\left(4\right)}^2-4\left(1\right)\left(-1\right)}}{2\left(1\right)}$

$$\begin{gathered} \Rightarrow m^2=\frac{-4\pm \sqrt{16+4}}{2} \\ \Rightarrow m^2=\frac{-4\pm 2\sqrt{5}}{2} \\ \Rightarrow m^2=-2\pm \sqrt{5} \end{gathered}$$

As the value of $m^2$ can not be negative, thus $m^2=-2+\sqrt{5}$

$\Rightarrow {\tan }^2\left(\frac{x}{2}\right)=-2+\sqrt{5}$

Hence, (D) is the correct option.