Triognometric identities?

Question

prove the following identities

3- 6 cos^2x

----------------- = 3(sin x +cos x)

sin x - cos x

Answer 1

L.H.S.=

3(1-2cos^2x)/(sinx-cosx)=

3cos2x(cosx+sinx)/(cos^2x-sin^2x)=

3cos2x(cosx+sinx)/cos2x)=

3(sinx+cosx)=

R.H.S.

Answer 2

LHS = (3 - 3 cos^2 x- 3 cos^2 x ) / (sin x - cos x)

={ 3(1-cos^2 x) - 3 cos ^2 x } / (sin x - cos x)

= (3 sin ^2 x - 3 cos ^2 x ) / (sin x - cos x)

= 3(sin x + cos x ) (sin x - cos x) / sin x -cos x)

= 3(sin x + cos x)

  

Answer 3

take 3 common from numerator u get

3(1-2cos^2x)          3cos2x

---------------------  =   ------------            

sinx -  cosx           sinx  -  cosx

using  cos2x=1-2cos^2x

=  3(cosx-sinx)(cosx + sinx)

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

     sinx  - cosx

so u get

    = 3(sinx + cosx)