How do you verify that #sin^2x/(cos^4x+cos^2xsin^2x)=tan^2x#?

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Bdub Share
Mar 8, 2018

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See Below

Explanation:

#LHS: sin^2x/(cos^4x+cos^2xsin^2x)#

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

#=sin^2x/(cos^2x*1)#->use property #sin^2x+cos^2x=1#

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

#=tan^2x#

#=RHS#

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