How do you prove #\frac { \tan v - \cot v } { \tan ^ { 2} v - \cot ^ { 2} v } = \sin v \cos v#?

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Answer 1 · 2017-09-25T05:37:45

Please see below.

#(tanv-cotv)/(tan^2v-cot^2v)#

= #((tanv-cotv))/((tanv+cotv)(tanv-cotv))#

= #1/(tanv+cotv)#

= #1/(sinv/cosv+cosv/sinv)#

= #1/((sin^v+cos^2v)/(sinvcosv)#

= #(sinvcosv)/(sin^v+cos^2v)#

= #(sinvcosv)/1#

= #sinvcosv#