If $\tan (\frac{\theta }{2})=\csc (\theta )-\sin (\theta )$, then the value of ${\tan }^2(\frac{\theta }{2})$ is

Solution

1. Given the equation $\tan (\frac{\theta }{2})=\csc (\theta )-\sin (\theta )$, we simplify the right side to $\frac{1-{\sin }^2(\theta )}{\sin (\theta )}=\frac{{\cos }^2(\theta )}{\sin (\theta )}$.
2. Using the half-angle substitutions $t=\tan (\frac{\theta }{2})$, $\sin (\theta )=\frac{2t}{1+t^2}$, and $\cos (\theta )=\frac{1-t^2}{1+t^2}$, the equation becomes $$\begin{array}{rl}t& =\frac{(1-t^2{)}^2}{(1+t^2{)}^2}\cdot \frac{1+t^2}{2t}\\ & =\frac{(1-t^2{)}^2}{2t(1+t^2)}\end{array}$$.
3. Rearranging the equation yields $2t^2(1+t^2)=(1-t^2{)}^2$, which simplifies to $2t^2+2t^4=1-2t^2+t^4$, or $t^4+4t^2-1=0$.
4. Applying the quadratic formula to $t^2$ gives $$\begin{array}{rl}{t}^2& =\frac{-4\pm \sqrt{16-4(1)(-1)}}{2}\\ & =-2\pm \sqrt{5}\end{array}$$.
5. Since $t^2={\tan }^2(\frac{\theta }{2})$ must be positive, we take the positive root $t^2=\sqrt{5}-2$.
6. Hence the answer is (A).