Simplify P(\sec \theta - \tan \theta)(1+\sin \theta)

Simplify $$\displaystyle{\left({\sec{\theta}}-{\tan{\theta}}\right)}{\left({1}+{\sin{\theta}}\right)}$$

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$$\displaystyle{\left({\sec{\theta}}-{\tan{\theta}}\right)}{\left({1}+{\sin{\theta}}\right)}=$$
$$\displaystyle={\sec{{\left(\theta\right)}}}{\sin{{\left(\theta\right)}}}-{\tan{{\left(\theta\right)}}}-{\tan{{\left(\theta\right)}}}{\sin{{\left(\theta\right)}}}=$$
$$\displaystyle={\frac{{{1}}}{{{\cos{\theta}}}}}+{\frac{{{\sin{\theta}}}}{{{\cos{\theta}}}}}-{\tan{\theta}}-{\frac{{{\sin{\theta}}}}{{{\cos{\theta}}}}}{\sin{\theta}}=$$
$$\displaystyle{\left[\because{\sec{\theta}}={\frac{{{1}}}{{{\cos{\theta}}}}}{\quad\text{and}\quad}{\tan{\theta}}={\frac{{{\sin{\theta}}}}{{{\cos{\theta}}}}}\right]}$$
$$\displaystyle={\frac{{{1}}}{{{\cos{\theta}}}}}+{\tan{\theta}}-{\tan{\theta}}-{\frac{{{{\sin}^{{2}}\theta}}}{{{\cos{\theta}}}}}=$$
$$\displaystyle={\frac{{{1}}}{{{\cos{\theta}}}}}-{\frac{{{{\sin}^{{2}}\theta}}}{{{\cos{\theta}}}}}=$$
$$\displaystyle={\frac{{{1}-{{\sin}^{{2}}\theta}}}{{{\cos{\theta}}}}}=$$
$$\displaystyle={\frac{{{{\cos}^{{2}}\theta}}}{{{\cos{\theta}}}}}={\cos{\theta}}$$
$$\displaystyle{\left[\because{{\sin}^{{2}}\theta}+{{\cos}^{{2}}\theta}={1}\right]}$$
Answer: $$\displaystyle{\left({\sec{\theta}}-{\tan{\theta}}\right)}{\left({1}+{\sin{\theta}}\right)}={\cos{\theta}}$$