Prove that #(sin3x)/cosx+(cos3x)/sinx=cotx-tanx# ? Please solve this help needed

1 Answer
Mar 19, 2018

See the proof below

Explanation:

We need

#cos(A-B)=cosAcosB+sinAsinB#

#cos2x=cos^2x-sin^2x#

#tanx=sinx/cosx#

#cotx=cosx/sinx#

Start with the #LHS# by placing on the same denominator

Therefore,

#LHS=(sin3x)/cosx+(cos3x)/sinx#

#=(sin3xsinx+cos3xcosx)/(cosxsinx)#

#=(cos3xcosx+sin3xsinx)/(cosxsinx)#

#=(cos(3x-x))/(cosxsinx)#

#=(cos2x)/(cosxsinx)#

#=(cosxcosx-sinxsinx)/((cosxsinx))#

#=(cosxcosx)/(cosxsinx)-(sinxsinx)/(cosxsinx)#

#=cosx/sinx-sinx/cosx#

#=cotx-tanx#

#=RHS#

#QED#