Question #7efd9

Question

1285 views around the world

You can reuse this answer
Creative Commons License

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