How do you simplify the following?

Question

$\frac{1 - {sin}^{2} x}{sin x + 1}$ $(1-sin^2 x)/(sin x + 1)$

$(tan x) (1 - {sin}^{2} x)$ $(tan x)(1-sin^2 x)$

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Answer 1 · 2016-03-26T21:21:56

$\frac{1 - {sin}^{2} x}{sin x + 1} = 1 - sin x$ $(1-sin^2 x)/(sin x + 1) = 1 - sin x$ when $x \neq \frac{3 π}{2} + 2 k π$ $x != (3pi)/2 + 2kpi$
$(tan x) (1 - {sin}^{2} x) = \frac{1}{2} sin 2 x$ $(tan x)(1 - sin^2 x) = 1/2 sin 2x$ when $x \neq k π$ $x != kpi$

Example 1.

Use the difference of squares identity:

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

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

$(1-sin^2 x)/(sin x + 1) = ((1-sin x)color(red)(cancel(color(black)((1+sin x)))))/color(red)(cancel(color(black)((1+sin x)))) = 1 - sin x$

with exclusion $sin x != -1$ (i.e. $x != (3pi)/2+2kpi$ )

Example 2.

Use the following:

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

$tan x = (sin x)/(cos x)$

$sin 2x = 2 sin x cos x$

as follows:

$(tan x)(1 - sin^2 x) =(sin x)/(cos x)*cos^2 x = sin x cos x = 1/2 sin 2x$

with exclusion $cos x != 0$ , i.e. $x != kpi$