How can $(sin^2(-x))/(tan^2(x))$ equal to $cos^2x$ ?

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Answer 1 · 2018-08-09T05:23:58

Please see below.

Here ,

$sin^2(-x)/tan^2x=[sin(-x)]^2/[sinx/cosx]^2=[-sinx]^2/((sinx)^2/(cosx)^2)$

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

Note that ,

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

Answer 2 · 2018-08-09T05:25:24

Please see The 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!

How can (sin^2(-x))/(tan^2(x))(sin^2(-x))/(tan^2(x)) equal to cos^2xcos^2x?