# Question #dd425

Jul 13, 2017

$\frac{{\sin}^{2} x \cos x}{1 + \sin x} ^ 2.$

#### Explanation:

The Expression =$\left(\frac{1}{\sec x + \tan x}\right) \left(\sin \frac{x}{\csc x + 1}\right) ,$

$= \left\{\frac{1}{\frac{1}{\cos} x + \sin \frac{x}{\cos} x}\right\} \left\{\sin \frac{x}{\frac{1}{\sin} x + 1}\right\} ,$

$= \left[\frac{1}{\frac{1 + \sin x}{\cos} x}\right] \left[\sin \frac{x}{\frac{1 + \sin x}{\sin} x}\right] ,$

$= \left\{\cos \frac{x}{1 + \sin x}\right\} \left\{{\sin}^{2} \frac{x}{1 + \sin x}\right\} ,$

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

Jul 13, 2017

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

#### Explanation:

We have: $\left(\frac{1}{\sec \left(x\right) + \tan \left(x\right)}\right) \left(\frac{\sin \left(x\right)}{\csc \left(x\right) + 1}\right)$

Let's apply three standard trigonometric identities; $\sec \left(x\right) = \frac{1}{\cos \left(x\right)}$, $\csc \left(x\right) = \frac{1}{\sin \left(x\right)}$ and $\tan \left(x\right) = \frac{\sin \left(x\right)}{\cos \left(x\right)}$:

$= \left(\frac{1}{\frac{1}{\cos \left(x\right)} + \frac{\sin \left(x\right)}{\cos \left(x\right)}}\right) \left(\frac{\sin \left(x\right)}{\frac{1}{\sin \left(x\right)} + 1}\right)$

$= \left(\frac{1}{\frac{1 + \sin \left(x\right)}{\cos \left(x\right)}}\right) \left(\frac{\sin \left(x\right)}{\frac{1 + \sin \left(x\right)}{\sin \left(x\right)}}\right)$

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

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