# (cos^2x-sin^2x)/cos^2x+sinxcosx=1-tanx

• Apr 6th 2009, 07:57 AM
skeske1234
(cos^2x-sin^2x)/cos^2x+sinxcosx=1-tanx
Can someone PLEASEEEEEE help me prove this trig identity.

I did it down to where I have
LS:
sinx/cosx

RS:
-sinx

I am not sure where I went wrong,
• Apr 6th 2009, 08:06 AM
LHS:
For numerator use formula

a^2 - b^2 = (a-b)(a+b)

So

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

Denominator

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

Divide numerator by denominator

The things in bold get cancelled

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One more thing try to avoid writing like thissssssss (Smile)

• Apr 6th 2009, 08:09 AM
stapel
Until you show your steps, I'm afraid we can't be sure where you went wrong, either. Sorry! (Blush)

• Apr 6th 2009, 08:30 AM
(Wait) Sorry to disappoint you SHOULD start a new thread for a new problem

Take Care of Yourself & Rules
• Apr 6th 2009, 09:19 AM
thepd
Quote:

LHS:
For numerator use formula

a^2 - b^2 = (a-b)(a+b)

So

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

Denominator

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

Divide numerator by denominator

The things in bold get cancelled

-------------------------------------

One more thing try to avoid writing like thissssssss (Smile)