How do you prove the identity #(sinx - cosx)/(sinx + cosx) = (2sin^2x-1)/(1+2sinxcosx)#?

Question

56905 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-09T19:46:18

See Below

Use the Property : #color(blue)(sin^2x+cos^2x=1#

#LHS : (sinx-cosx)/(sinx+cosx)#

#=(sinx-cosx)/(sinx+cosx)* (sinx+cosx)/(sinx+cosx)#-> multiply by conjugate

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

#=(sin^2x-[1-sin^2x])/([sin^2x+cos^2x]+2sinxcosx)#

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

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

#=RHS#