Question

Show that cos^2/1- tan + sin^3/ sin - cos = 1+ sin , cos

Answer 1

hope it will help you

Answer 2

cos^2/1- tan + sin^3/ sin - cos = 1+ sin , cos

Proved below.

Step-by-step explanation:

Given:

Let $\frac{cos^2x}{1 - tan^2x} +\frac{sin^3x}{Sin x - Cos x} =1 +sinx cosx$

L.H.S.

=$\frac{cos^2x}{1 - tan^2x} +\frac{sin^3x}{sin x - cos x}$

= $\frac{cos^2x}{1-\frac{sinx}{cosx}}+\frac{sin^3x}{sinx-cosx}$  [$tanx=\frac{sinx}{cosx}$]        

= $\frac{cos^2x}{\frac{cosx-sinx}{cosx}}+\frac{sin^3x}{sinx-cosx}$

= $\frac{cos^3x}{cosx-sinx}+\frac{sin^3x}{sinx-cosx}$

=$\frac{cos^3x}{-(sinx-cosx)}+\frac{sin^3x}{sinx-cosx}$

=$\frac{sin^3x-cos^3x}{sinx-cosx}$

=$\frac{(sinx - cosx)(sin^2x + cos^2x +sin x cos x)}{sinx-cosx}$   [$a^3 - b^3 = ( a - b )(a^2 + b^2 + ab )$]

=$(sin ^2 x + cos^2 x +sin x cos x )$

=$1 + sin xcos x$                           [$sin ^2 x + cos^ 2 x = 1$]

=R.H.S.

Hence proved.