How can (sin^2(-x))/(tan^2(x))sin2(x)tan2(x) equal to cos^2xcos2x?

2 Answers
Aug 9, 2018

Please see below.

Explanation:

Here ,

sin^2(-x)/tan^2x=[sin(-x)]^2/[sinx/cosx]^2=[-sinx]^2/((sinx)^2/(cosx)^2)sin2(x)tan2x=[sin(x)]2[sinxcosx]2=[sinx]2(sinx)2(cosx)2

:.sin^2(-x)/tan^2x=(sinx)^2/(sinx)^2xx(cosx)^2=(cosx)^2=cos^2x

Note that ,

sin(-x)/tanx=(-sinx)/(sinx/cosx)=-sinx/sinx xxcosx=-cosx

Aug 9, 2018

Please see The Explanation.

Explanation:

Recall that, sin(-x)=-sinx.

:. sin^2(-x)={sin(-x)}^2=(-sinx)^2=sin^2x.

"Hence, the Expression"=sin^2(-x)/tan^2x,

=sin^2x-:tan^2x,

=sin^2x-:sin^2x/cos^2x,

=(cancel(sin^2x))(cos^2x/cancel(sin^2x)),

=cos^2x, as desired!