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!