# If $\tan{\frac{x}{2}}=\csc x - \sin x$, then find the value of $\tan^2{\frac{x}{2}}$.

If $$\tan{\frac{x}{2}}=\csc x - \sin x$$, then find the value of $$\tan^2{\frac{x}{2}}$$.

HINT: The answer is $$-2\pm \sqrt5$$.

What I have tried so far: $$\tan{\frac{x}{2}} = \frac{1}{\sin x}-\sin x$$ $$\tan{\frac{x}{2}} = \frac{1-\sin^2 x}{\sin x}$$ $$\tan{\frac{x}{2}} = \frac{\cos^2 x}{\sin x}$$

I don't know how to solve this problem. Pls help. Thank you :)

• Let $u=\frac x2$. – Szeto Feb 14 at 9:49

## 3 Answers

Notice that $$\cos(x) = \frac{1 -\tan^2 (x/2)}{1+\tan^2(x/2)}$$ And that $$\tan^2(x/2) = \frac{\cos^4(x)}{1-\cos^2(x)}$$ Let $$t = \cos(x)$$. Plug the 2nd equation into the first one and after some algebra, we get $$(1-t)(1-t^2-t^4) = 0$$

$$\cos^2x=\tan\dfrac x2\sin x=\dfrac{\sin\dfrac x2}{\cos\dfrac x2}\cdot2\sin\dfrac x2\cos\dfrac x2=2\sin^2\dfrac x2=1-\cos x$$

$$\implies\cos x=\dfrac{-1\pm\sqrt5}2$$

As for real $$x,\cos x\ge-1,\cos x\ne\dfrac{-1-\sqrt5}2<-1$$

Using Weierstrass Substitution $$\dfrac{1-\tan^2\dfrac x2}{1+\tan^2\dfrac x2}=\cos x=\dfrac{-1+\sqrt5}2$$

Now apply Componendo et Dividendo

• Please find the updated answer. Please feel free to pinpoint any mistake/doubt – lab bhattacharjee Feb 15 at 5:57

Hint: Use that $$\frac{1}{\sin(x)}-\sin(x)=1/2\,{\frac {1+ \left( \tan \left( x/2 \right) \right) ^{2}}{\tan \left( x/2 \right) }}-2\,{\frac {\tan \left( x/2 \right) }{1+ \left( \tan \left( x/2 \right) \right) ^{2}}}$$