Question #7efd9

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Answer 1 · 2017-02-07T17:29:16

see below

Left Hand Side:

$(tanx-cotx)/(tan^2x-cot^2x) = (tanx-cot x)/((tanx-cotx)(tanx+cotx))$

$=(cancel(tanx-cot x) )/(cancel((tanx-cotx))(tanx+cotx))$

$=1/(tanx+cotx)$

$=1/(sinx/cos x +cosx/sinx)$

$=1/((sin^2x+cos^2x)/(sinx cosx)$

$=1/(1/(sinx cosx))$

$=1* (sinxcosx)/1$

$=sinxcosx$

$:.=$ Right Hand Side