Question

(1+cotx+tanx)(sinx-cosx)bysec^3x-cosec^3x=sin^2xcos^2x

Answer 1

=(sinx.cosx + 1)/(1+cosx.sinx)×(sinx.cosx)
=sinx.cosx
=R.H.S
Hence,Proved.
I have used the identity.
$x {}^{3} - y {}^{3} = (x - y)( {x}^{2} + xy + y {}^{2} )$

Answer 2

(1+cotx+tanx)(sinx-cosx)bysec^3x-cosec^3x=sin^2xcos^2x

Proved below.

Step-by-step explanation:

Given:

$\frac{(1+cotx+tanx)(sinx-cosx)}{sec^{3}x-cosec^3x} =sec^2xcos^2x$

L.H.S=$\frac{(1+cotx+tanx)(sinx-cosx)}{sec^{3}x-cosec^3x}$

We know that,

$cosecx=\frac{1}{x}$ ; $sec x= \frac{1}{cos x}$; $tan x=\frac{sin x}{cos x}$; $cot x= \frac{cos x}{sinx}$

=$\frac{(1+\frac{cosx}{sinx}+\frac{sinx}{cosx})(sinx-cosx)}{\frac{1}{cos^3x}-\frac{1}{sin^3x}}$

=$[tex]\frac{(\frac{sinxcosx+cos^2x+sin^2x}{(sinxcosx)}) (sinx-cosx)}{\frac{sin^3x-cos^3x}{sin^3xcos^3x} }$[/tex]

=$\frac{(sinxcosx+cos^2x+sin2x)(sinx-cosx)(sin^2xcos^2x)}{sin^3x-cos^3x}$

We know that

$a^3 - b^3 = (a - b)(a^2 + b^2+ ab)$

Also $sin^2x+cos^2x=1$

= $\frac{(1+sinxcosx)(sin-cosx)(sin^2xcos^2x)}{(sinx-cosx)(sin^2x+cos^2x+sinxcosx)}$

=$\frac{(1+sinxcosx)(sin-cosx)(sin^2xcos^2x)}{(sinx-cosx)(1+sinxcosx)}$

=sin^2xcos^2x

=R.H.S.

Hence proved.