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

1. ## (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,
can someone please help

2. 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

Adarsh

3. Until you show your steps, I'm afraid we can't be sure where you went wrong, either. Sorry!

Please reply with a clear listing of your work and reasoning so far. Thank you!

4. Sorry to disappoint you SHOULD start a new thread for a new problem

Take Care of Yourself & Rules
Adarsh

5. Originally Posted by ADARSH
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

Adarsh
.....ok that's fine. Now just divide the Numerator by denominator...
and the answer comes ..........(cos(x)-sin(x))/cos(x)=1- tan(x);
that's it. what is the problem?