Question #5abd9

Question

990 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-14T02:26:34

The right side of the equation was missing a minus sign between the #cotx# and #tanx#.

#(1-2(sinx)^2)/((sinx)(cosx))#

#=1/((sinx)(cosx))-(2(sinx)^2)/((sinx)(cosx))#

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

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

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

#=tanx+cotx-2tanx#

#=cotx-tanx#

Answer 2 · 2017-11-14T06:21:33

#LHS=(1-2sin^2x)/(sinxcosx)#

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

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

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

#=cosx/sinx-sinx/cosx#

#=cotx-tanx#