Thread: (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

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

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!

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

Take Care of Yourself & Rules
Adarsh

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?