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,

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

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

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

Take Care of Yourself & Rules

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

.....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?

,

,

,

### cos x / 1-tanx - sin^2x/cos x - sinx =cosx sinx

Click on a term to search for related topics.