How do you prove #cot^2x - cos^2x = cot^2cos^2#?

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Answer 1 · 2017-05-26T08:28:52

see below

to prove

#cot^2x-cos^2x=cot^2xcos^2x#

take LHS and change to cosines an sines and then rearrange to arrive at the RHS

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

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

factorise numerator

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

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

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

#=cos^2xcot^2x=cot^2xcos^2x#

=RHS as reqd.