Question #de16d

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Answer 1 · 2016-08-06T08:02:48

We have: #(cos^2x-sin^2x)/(cot^2x-tan^2x)#

Start by writing tan and cot in terms of sin and cos

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

#=(cos^2x-sin^2x)/((cos^4x-sin^4x)/(sin^2xcos^2x))*(sin^2xcos^2x)/(sin^2xcos^2x)#

#=(cos^4xsin^2x-sin^4xcos^2x)/(cos^4x-sin^4x)#

#=((sin^2xcos^2x)cancel((cos^2x-sin^2x)))/((cos^2x+sin^2x)cancel((cos^2x-sin^2x)))#

#=(sin^2xcos^2x)/1#

#=sin^2xcos^2x#

Answer 2 · 2016-08-06T09:29:02

We have:
#LHS=(cos^2x-sin^2x)/(cot^2x-tan^2#
#=(sin^2xcos^2x(cos^2x/(sin^2xcos^2x)-sin^2x/(sin^2xcos^2x)))/(csc^2x-1-sec^2x+1)#

#=((sin^2xcos^2x)(csc^2x-sec^2x))/(csc^2x-sec^2x)#

=#sin^2xcos^2x#