algebra precalculus - Prove that $frac{2sin(a)+sec(a)}{1+tan(a)}$ = $frac{1+tan(a)}{sec(a)}$

Prove that $\frac{2\sin(a)+\sec(a)}{1+\tan(a)}$ = $\frac{1+\tan(a)}{\sec(a)}$

My attempt using the LHS

$$\frac{2\sin(a)+\sec(a)}{1+\tan(a)}$$

$$\frac{2\sin(a)+\frac{1}{\cos(a)}}{1+\frac{\sin(a)}{\cos(a)}}$$

$$\frac{\frac{2\sin(a)+1}{\cos{a}}}{\frac{\cos(a)+\sin(a)}{\cos(a)}}$$

$${\frac{2\sin(a)+1}{\cos{a}}} * {\frac{\cos(a)}{\cos(a)+sin(a)}}$$

$$\frac{2\sin(a)+1}{\cos(a)+sin(a)}$$

Now I am stuck...

Answer

Hint: What does $(\cos a+ \sin a)^2$ simplify to?