Prove $\frac{\sin\theta}{1+\cos\theta} + \frac{1+\cos\theta}{\sin\theta} = \frac{2}{\sin\theta}$ How to prove:
$\frac{\sin\theta}{1+\cos\theta} + \frac{1+\cos\theta}{\sin\theta} = \frac{2}{\sin\theta}$
Please help.
 A: $$\dfrac{\sin(\theta)}{1+\cos(\theta)}+\dfrac{1+\cos(\theta)}{\sin(\theta)}=\dfrac{\sin^2(\theta)+(1+\cos(\theta))^2}{\sin(\theta)(1+\cos(\theta))}$$
$$=\dfrac{\sin^2(\theta)+\cos^2(\theta)+1+2\cos(\theta)}{\sin(\theta)(1+\cos(\theta))}$$
$$=\dfrac{2+2\cos(\theta)}{\sin(\theta)(1+\cos(\theta))}$$
$$=\dfrac{2(1+\cos(\theta))}{\sin(\theta)(1+\cos(\theta))}$$
$$=\dfrac{2}{\sin(\theta)}$$
A: Common denominator:
$$\frac{\sin^2\theta+1+\cos^2\theta+2\cos\theta}{\sin\theta(1+\cos\theta)}$$
$$=\frac{2(1+\cos\theta)}{\sin\theta(1+\cos\theta)}=\frac{2}{\sin\theta}$$
A: Let $u=\theta/2$. Then, by using double-angle formulas, $\sin \theta = 2 \cos u \sin u$, $1+\cos \theta = 2\cos^2 u$.
So the left-hand side is equal to
$$
\frac{2\cos u \sin u}{2\cos^2 u} + \frac{2 \cos^2 u}{2\cos u \sin u} = \frac{\cos u}{\sin u} + \frac{\sin u}{\cos u} = \frac{\cos^2 u + \sin^2 u}{\cos u \sin u}=\frac{1}{\cos u \sin u} \, ,
$$
while the right-hand side is equal to
$$
\frac{2}{2 \cos u \sin u}=\frac{1}{\cos u \sin u} \, .
$$
A: $$\sin^2\theta=1-\cos^2\theta=(1-\cos\theta)(1+\cos\theta)$$
$$\implies \frac{\sin\theta}{1+\cos\theta}=\frac{1-\cos\theta}{\sin\theta}$$
Now add the second term
