How do you simplify the following?

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

Mar 26, 2016
1. $\frac{1 - {\sin}^{2} x}{\sin x + 1} = 1 - \sin x$ when $x \ne \frac{3 \pi}{2} + 2 k \pi$

2. $\left(\tan x\right) \left(1 - {\sin}^{2} x\right) = \frac{1}{2} \sin 2 x$ when $x \ne k \pi$

Explanation:

Example 1.

Use the difference of squares identity:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

with $a = 1$, $b = \sin x$ as follows:

$\frac{1 - {\sin}^{2} x}{\sin x + 1} = \frac{\left(1 - \sin x\right) \textcolor{red}{\cancel{\textcolor{b l a c k}{\left(1 + \sin x\right)}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{\left(1 + \sin x\right)}}}} = 1 - \sin x$

with exclusion $\sin x \ne - 1$ (i.e. $x \ne \frac{3 \pi}{2} + 2 k \pi$)

Example 2.

Use the following:

${\sin}^{2} x + {\cos}^{2} x = 1$ in the form $1 - {\sin}^{2} x = {\cos}^{2} x$

$\tan x = \frac{\sin x}{\cos x}$

$\sin 2 x = 2 \sin x \cos x$

as follows:

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

with exclusion $\cos x \ne 0$, i.e. $x \ne k \pi$