Question

Prove that? `(cos^4 x - sin^4 x)/(1-tan x) -= cos^2 x +sin x cos x`

Answer 1

Please see proof below.

Explanation:

`(cos^4 x - sin^4 x)/(1-tan x) = cos^2 x +sin x cos x`

`LHS=(cos^4 x - sin^4 x)/(1-tan x)`

`color(white)(LHS)=((cos^2x-sin^2x)(cos^2x+sin^2x))/(1-tanx)`

`color(white)(LHS)=((cos^2x-sin^2x)*1)/(1-tanx)`

`color(white)(LHS)=(cos^2x-sin^2x)/(1-tanx)`

`color(white)(LHS)=((1-sin^2x)-sin^2x)/(1-tanx)`

`color(white)(LHS)=(1-2sin^2x)/(1-tanx)`

`color(white)(LHS)=((1-2sin^2x)(1+tanx))/((1-tanx)(1+tanx))`

`color(white)(LHS)=((1-2sin^2x)(1+tanx))/(1-tan^2x)`

`color(white)(LHS)=((1-2sin^2x)(1+tanx))/(cos^2x/cos^2x-sin^2x/cos^2x)`

`color(white)(LHS)=((1-2sin^2x)(1+tanx))/((cos^2x-sin^2x)/cos^2x)`

`color(white)(LHS)=((1-2sin^2x)(1+tanx))/(((1-sin^2x)-sin^2x)/cos^2x)`

`color(white)(LHS)=((1-2sin^2x)(1+tanx))/((1-2sin^2x)/cos^2x)`

`color(white)(LHS)=(1-2sin^2x)(1+tanx)*(cos^2x/(1-2sin^2x))`

`color(white)(LHS)=color(red)(cancel(color(black)((1-2sin^2x))))(1+tanx)*cos^2x/color(red)(cancel(color(black)((1-2sin^2x))`

`color(white)(LHS)=(1+tanx)*cos^2x`

`color(white)(LHS)=cos^2x+cos^2x*tanx`

`color(white)(LHS)=cos^2x+cos^2x*sinx/cosx`

`color(white)(LHS)=cos^2x+cos^color(red)(cancel(color(black)(2)))x*sinx/color(red)(cancel(color(black)(cosx)))`

`color(white)(LHS)=cos^2x+cosxsinx`

`color(white)(LHS)=cos^2x+sinxcosx`

`color(white)(LHS)=RHS`

Answer 2

We seek to prove that the following is an indetiuty:

`(cos^4 x - sin^4 x)/(1-tan x) -= cos^2 x +sin x cos x`

Consider the LHS::

`LHS -= (cos^4 x - sin^4 x)/(1-tan x)`
`\ \ \ \ \ \ \ \ = ((cos^2 x)^2 - (sin^2 x)^2)/(1-tan x)`

This is the difference of two squares, so we can use the identity:

`A^2 - B^2 -= (A+B)(A-B)`

Thus:

`LHS = ((cos^2 x+sin^2 x)(cos^2 x-sin^2 x))/(1-tan x)`

Now, And again, we have the difference of two squares, so we can factor further, and also we use the trigonometric identity:

`sin^2x+cos^2x -= 1`

Thus:

`LHS = ((1)(cosx+sinx)(cos x-sin x))/(1-tan x)`

`\ \ \ \ \ \ \ \ = ((cosx+sinx)(cos x-sin x))/(1-sinx/cosx)`

`\ \ \ \ \ \ \ \ = ((cosx+sinx)(cos x-sin x))/((cosx-sinx)/cosx)`

`\ \ \ \ \ \ \ \ = ((cosx+sinx)(cos x-sin x)cosx)/((cosx-sinx))`

`\ \ \ \ \ \ \ \ = (cosx+sinx)cosx`

`\ \ \ \ \ \ \ \ = cos^2x+sinxcosx`

`\ \ \ \ \ \ \ \ = RHS \ \ \ \` QED