Answer
$\frac{sec~\theta(1+tan~\theta)}{sec~\theta+csc~\theta}=\frac{sin~\theta}{cos~\theta}$
$ tan \theta $
Work Step by Step
$\frac{sec~\theta(1+tan~\theta)}{sec~\theta+csc~\theta}=\frac{\frac{1}{cos~\theta}(1+\frac{sin~\theta}{cos~\theta})}{\frac{1}{cos~\theta}+\frac{1}{sin~\theta}}=\frac{\frac{1}{cos~\theta}+\frac{sin~\theta}{cos^2\theta}}{\frac{sin~\theta+cos~\theta}{cos~\theta~sin~\theta}}=\frac{\frac{cos~\theta+sin~\theta}{cos^2\theta}}{\frac{sin~\theta+cos~\theta}{cos~\theta~sin~\theta}}=\frac{cos~\theta+sin~\theta}{cos^2\theta}~\frac{cos~\theta~sin~\theta}{cos~\theta+sin~\theta}=\frac{sin~\theta}{cos~\theta}$
