How to prove this trigonometric identity?

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1 Answer

Answer:

See below

Explanation:

#cos2x=cos^2x-sin^2#
#sin2x=2sinxcosx#
Are you happy about these identities?
If so then proving identities, as above, is a matter of doing something and seeing where you get to!!!
LHS can be written

#(cosx-(cos^2x-sin^2x)+2)/(3sinx-2sinxcosx)#
=#(cosx-cos^2x+sin^2x+2)/(sinx(3-2cosx))#
RHS #(1+cosx)/sinx#
Multiply LHS and RHS by #sinx(3-2cosx)#
LHS=#cosx-cos^2x+sin^2x+2#

RHS=#(1+cosx)(3-2cosx)#
=#3-2cosx+3cosx-2cos^2x#
=#3+cosx-2cos^2x#
Now looking at the LHS and remembering that #sin^2x+cos^2x=1#or # sin^2x=1-cos^2x#
LHS=#cosx-cos^2x+(1-cos^2x)+2#
=#cosx-2cos^2x+3#

We are there!!!
Difficult for me to write this really clearly here, paper would be easier!!!
BUT THE IMPORTANT THING WITH IDENTITIES IS DO SOMETHING AND YOU WILL GET THERE.