Question #6ef07

3 Answers
Jan 24, 2018

The identity is not valid.

Explanation:

The identity is not valid.

In order for an identity to be valid,
it have to work for all values of the variable

For a counterexample choose #x=pi/2#

#(1-2sin(pi/2))/(cos(pi/2)-sin(pi/2))=cos(pi/2)-sin(pi/2)#

#1=-1#

Which obviously isn't true

Jan 24, 2018

I tried this:

Explanation:

Have a look:

enter image source here

Jan 24, 2018

The identity is true if there is #2sinx*cosx# in the numerator of #LHS#.

Explanation:

#LHS=(1-2sinx*cosx)/(cosx-sinx)#

#=(sin^2x-2sinx*cosx+cos^2x)/(cosx-sinx)#

#=(cosx-sinx)^2/(cosx-sinx)#

#=cosx-sinx=RHS#