#(1+cotx)=(2cos2x)/(sin2x-2sin^2x)# prove?

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Answer 1 · 2018-06-04T07:00:17

#RHS=(2cos2x)/(sin2x-2sin^2x)#

#=(2(cos^2x-sin^2x))/(2sinx*cosx-2sin^2x)#

#=(cancel(2)(cosx+sinx)cancel((cosx-sinx)))/(cancel(2)sinxcancel((cosx-sinx)))#

#=(cosx+sinx)/sinx=cosx/sinx+sinx/sinx=1+cotx=LHS#

Answer 2 · 2018-06-04T07:01:59

#RHS=(2cos2x)/(sin2x-2sin^2x)#

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

#=(2(cosx-sinx)(cosx+sinx))/(2sinx(cosx-sinx))#

#=(cosx+sinx)/sinx#

#=cosx/sinx+sinx/sinx#

#=1+cotx=LHS#