Prove the given identity: $\frac{{1-\cos x}}{1 + \cos x}=\tan^2\frac{{x}}{2}$

Question

$$\frac{{1-\cos x}}{1 + \cos x}=\tan^2\frac{{x}}{2}$$

I've struggled with this one for a while now. If anyone can break it down in as many steps as possible that'd be great. Also, what are ways of telling which side might be easier to pick for doing the proof?

Answer 1

Given $$\dfrac{1-\cos x}{1+\cos x}=\tan^2\left(\dfrac x2\right)$$

Now take $u=\dfrac x2$ then we get, $$\dfrac{1-\cos(2u)}{1+\cos(2u)}=\tan(u)$$

$$\mbox{R.H.S}=\tan^2(u)$$ $$=\left(\dfrac{\sin(u)}{\cos(u)}\right)^2=\dfrac{\sin^2u}{\cos^2u}=\dfrac{(1-\cos(2u))\cdot\dfrac12}{(1+\cos(2u))\cdot\dfrac12}=\dfrac{1-\cos(2u)}{1+\cos(2u)}=\mbox{L.H.S}$$

Therefore, $$\dfrac{1-\cos x}{1+\cos x}=\tan^2\left(\dfrac x2\right)$$

Answer 2

hint Use the identity

$$\cos(2t)=1-2\sin^2(t)=2\cos^2(t)-1$$

Take $t=\frac x2$.

Answer 3

Recall that by Tangent half-angle substitution

$$\cos x = \frac{1-\tan^2 \frac x 2}{1+\tan^2 \frac x 2}\implies 1-\cos x=\frac{2\tan^2 \frac x 2}{1+\tan^2 \frac x 2} \quad 1+\cos x=\frac{2}{1+\tan^2 \frac x 2}$$

and therefore

$$\frac{{1-\cos x}}{1 + \cos x}=\frac{2\tan^2 \frac x 2}{1+\tan^2 \frac x 2}\frac{1+\tan^2 \frac x 2}2=\tan^2\frac{{x}}{2}$$

Answer 4

Note that $$\cos x =1-2 \sin ^2 (x/2)$$

Thus $$\frac{ {1-\cos x}}{1 + \cos x}= \frac {2 \sin ^2 (x/2)}{ 2-2 \sin ^2 (x/2) }= \frac { \sin ^2 (x/2)}{ \cos ^2 (x/2) } = \tan^2\frac{{x}}{2}$$

Answer 5

The derivatives of both sides are the same. Now plug in $x=0$.