# Prove that cos x+sinx /cosx-sinx equal to 1+sin2x/cos2x

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The expression you have provided is true because sin 2x = 2*sin x*cos x and cos 2x = (cos x)^2 - (sin x)^2

Let start with (cos x + sin x)/(cos x - sin x)

multiply the numerator and denominator by (cos x + sin x)

=> (cos x + sin x)(cos x + sin x)/(cos x - sin x)(cos x + sin x)

=> (cos x + sin x)^2/(cos x)^2 - (sin x)^2

=> [(cos x)^2 + 2*sin x*cos x + (sin x)^2]/cos 2x

=> (1 + sin 2x)/cos 2x

**This proves that (cos x + sin x)/(cos x - sin x) = (1 + sin 2x)/cos 2x**

L:H:S ≡ (cosx+sinx)/(cosx-sinx)

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

= (cos²x + 2sinx.cosx + sin²x)/(cos²x-sin²x)

**we know that sin²x+cos²x=1, 2sinx.cosx = sin2x and cos²x-sin²x=cos2x**

= (1 + sin2x)/cos2x