Question

`(sinx-sinxcosx)/(sinx+sinxtanx)=(1-cosx)/(1+tanx)`?

Answer 1

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

Explanation:

We have:

`(sinx-sinxcosx)/(sinx+sinxtanx)=(1-cosx)/(1+tanx)`

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

`(sinx(1-cosx))/(sinx(1+tanx))=(1-cosx)/(1+tanx)`

which I can write as:

`(sinx/sinx)(1-cosx)/(1+tanx)=(1-cosx)/(1+tanx)`

`(1)(1-cosx)/(1+tanx)=(1-cosx)/(1+tanx)`

`(1-cosx)/(1+tanx)=(1-cosx)/(1+tanx)color(white)(000)color(green)sqrt`