How do I simplify this trigonometric expression?

Question

How do I simplify $$\frac{\sin(x)}{1+\cos(x)}+\frac{1+\cos(x)}{\sin(x)}?$$ Any help would be appreciated.

Answer 1

$$\frac{\sin(x)}{1+\cos(x)}+\frac{1+\cos(x)}{\sin(x)}=\frac{\sin^{2}(x)+\cos^{2}(x)+2\cos(x)+1}{\sin(x)(1+\cos(x))}=\frac{2}{\sin(x)}$$

Answer 2

This is not recommended as the best way here, but can work if stuck. Let $t=\tan \frac x2$ so that $$\sin x = \frac {2t}{1+t^2}; \cos x =\frac {1-t^2}{1+t^2}$$The original expression becomes (after multiplying numerators and denominators by $1+t^2$) $$\frac {2t}{1+t^2+1-t^2}+\frac {1+t^2+1-t^2}{2t}=t+\frac 1t=\frac {t^2+1}t=\frac 2{\sin x}$$

Answer 3

Using the identity $\frac{\sin(x)}{1+\cos(x)}=\frac{1-\cos(x)}{\sin(x)}$ and substituting in the given, we get

$$\frac{\sin(x)}{1+\cos(x)}+\frac{1+\cos(x)}{\sin(x)}=\frac{1-\cos(x)}{\sin(x)}+\frac{1+\cos(x)}{\sin(x)} =\frac{2}{\sin(x)}$$