Please prove this identity

Okay, let's start by multiplying the right hand side by $$\frac{1+cos(\theta)}{1+cos(\theta)}$$

We now have

$$\frac{sin(\theta)+tan(\theta)}{1+cos(\theta)} = \frac{tan(\theta) (1+cos(\theta))}{1+cos(\theta)}$$

Work out the numerator on the right hand side

$$\frac{sin(\theta)+tan(\theta)}{1+cos(\theta)} = \frac{tan(\theta) cos(\theta)+tan(\theta)}{1+cos(\theta)}$$

Use $$tan(\theta) = \frac{sin(\theta)}{cos(\theta)}$$

$$\frac{sin(\theta)+tan(\theta)}{1+cos(\theta)} = \frac{\frac{sin(\theta)}{cos(\theta)} cos(\theta)+tan(\theta)}{1+cos(\theta)}$$

and now we have

$$\frac{sin(\theta)+tan(\theta)}{1+cos(\theta)} = \frac{sin(\theta)+tan(\theta)}{1+cos(\theta)}$$

Reinout