How do you verify:#cot^2x-cos^2x=cot^2xcos^2x#?

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13
Mar 22, 2017

Answer:

Please see below.

Explanation:

#cot^2x-cos^2x#

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

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

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

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

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

= #cot^2xcos^2x#

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Nghi N. Share
Mar 22, 2017

Develop the left side:
#LS = (cos^2 x)/(sin^2 x) - cos ^2 x = ((cos^2 x)(1 - sin^2 x))/(sin^2 x) =#
#= (cos^2 x.cos^2 x)/(sin^2 x) = cot^2 x.cos^2 x# Proved.

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