How do you verify #sin^2 (-x) / tan^2 x = cos^2 x#?

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Answer 1 · 2016-04-16T22:05:10

We will be using the following:

Then, when #cos(x)!=0# and #sin(x)!=0#, we have

#sin^2(-x)/tan^2(x) = (sin(-x))^2/(sin(x)/cos(x))^2#

#=(-sin(x))^2/(sin^2(x)/cos^2(x))#

#=sin^2(x)/(sin^2(x)/cos^2(x))#

#=1/(1/cos^2(x))#

#=cos^2(x)#