How do you show that #cosx/(1 - tanx) - sin^2x/(cosx- sinx) = cosx + sinx#?

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Answer 1 · 2016-10-31T22:33:49

The following identities are important for this problem:

-#tantheta = sintheta/costheta#

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

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

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

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

Factor the numerator on the left as a difference of squares, #a^2 - b^2 = (a + b)(a - b)#.

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

#cosx + sinx = cosx + sinx#

Identity Proved!!

Hopefully this helps!

Answer 2 · 2016-10-31T22:36:23

#LHS=cosx/(1-tanx)-sin^2x/(cosx-sinx)#

#=cos^2x/(cosx(1-sinx/cosx))-sin^2x/(cosx-sinx)#

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

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

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

#=cosx+sinx=RHS#

Proved