$\frac{sin x - sin x cos x}{sin x + sin x tan x} = \frac{1 - cos x}{1 + tan x}$ $(sinx-sinxcosx)/(sinx+sinxtanx)=(1-cosx)/(1+tanx)$ ?

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Answer 1 · 2017-02-23T09:22:28

Factor out $sin x$ $sinx$ in the numerator and denominator of the left side fraction. See below for details:

We have:

$\frac{sin x - sin x cos x}{sin x + sin x tan x} = \frac{1 - cos x}{1 + tan x}$ $(sinx-sinxcosx)/(sinx+sinxtanx)=(1-cosx)/(1+tanx)$

I'm first going to factor out $sin x$ $sinx$ in the numerator and denominator on the left side:

$\frac{sin x (1 - cos x)}{sin x (1 + tan x)} = \frac{1 - cos x}{1 + tan x}$ $(sinx(1-cosx))/(sinx(1+tanx))=(1-cosx)/(1+tanx)$

which I can write as:

$(\frac{sin x}{sin x}) \frac{1 - cos x}{1 + tan x} = \frac{1 - cos x}{1 + tan x}$ $(sinx/sinx)(1-cosx)/(1+tanx)=(1-cosx)/(1+tanx)$

$(1) \frac{1 - cos x}{1 + tan x} = \frac{1 - cos x}{1 + tan x}$ $(1)(1-cosx)/(1+tanx)=(1-cosx)/(1+tanx)$

$\frac{1 - cos x}{1 + tan x} = \frac{1 - cos x}{1 + tan x} 000 \sqrt{}$ $(1-cosx)/(1+tanx)=(1-cosx)/(1+tanx)color(white)(000)color(green)sqrt$

sinx−sinxcosxsinx+sinxtanx=1−cosx1+tanx(sinx-sinxcosx)/(sinx+sinxtanx)=(1-cosx)/(1+tanx)?