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

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