# Proving Trigonometric identity

• Apr 26th 2011, 04:36 PM
Devi09
Proving Trigonometric identity
((cos^2)x - (sin^2)x) /( (cos ^2)x + SinxCosx) = 1 - tanx

I simplified the right side to (cosx - sinx) / (cosx)

Don't know what to do after
• Apr 26th 2011, 04:59 PM
convert

1 - tan x into it's parts

1 - tan x => 1 - sin x / cos x

(cos x - sin x) / (cos x) = 1 - sin x / cos x

multiply each side by cos x

(cos x) (cos x - sin x) / (cos x) = (cos x) (1 - sin x / cos x)

(cos x - sin x) = (cos x - sin x)

You should be able to take it from there.
• Apr 26th 2011, 07:11 PM
Soroban
Hello, Devi09!

Quote:

[(cos x)^2 - (sin x)^2] / [(cos x)^2 + (sin x)(cos x)] .= .1 - tan x

I simplified the right side to: .(cos x - sin x) / (cos x)

. . . . . . . . . . . . .(cos x)^2 - (sin x)^2 . . . . . (cos x - sin x)(cos x + sin x)
The left side: . ----------------------------- . = . -------------------------------
. . . . . . . . . . . (cos x)^2 + (sin x)(cos x) . . . . . . cos x(cos x + sin x)

. . . . . . cos x - sin x . . . . cos x . . sin x
Then: . -------------- . = . ------- - ------ . = . 1 - tan x
. . . . . . . . cos x . . . . . . . cos x . . cos x

• Apr 27th 2011, 07:14 PM
Devi09
Quote:

Originally Posted by Soroban
Hello, Devi09!

. . . . . . . . . . . . .(cos x)^2 - (sin x)^2 . . . . . (cos x - sin x)(cos x + sin x)
The left side: . ----------------------------- . = . -------------------------------
. . . . . . . . . . . (cos x)^2 + (sin x)(cos x) . . . . . . cos x(cos x + sin x)

. . . . . . cos x - sin x . . . . cos x . . sin x
Then: . -------------- . = . ------- - ------ . = . 1 - tan x
. . . . . . . . cos x . . . . . . . cos x . . cos x

For the denominator of the first line how does (cos x)^2 + (sin x)(cos x) become cos x(cos x + sin x) ?

and for the last line, how does (cos x - sin x)/(cos x) become (cosx / cosx) - (sinx /cosx)