How do you simplify the expression #(cot^2tcos^2t)/(cot^2t-cos^2t)#?

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Answer 1 · 2016-09-29T05:49:19

1

Since #cot t=cos t/sin t#, let's substitute to have:

#(cos^2 t/sin^2t cos^2t)/(cos^2 t/sin^2 t-cos^2 t)#

that's

#(cos^4t/cancel(sin^2t))/((cos^2t-sin^2tcos^2t)/cancel(sin^2t))#

#=cos^4t/(cos^2t(1-sin^2t)#

that's, since #1-sin^2t=cos^2t#,

#cos^4t/cos^4t=1#