Question #6ef07

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Answer 1 · 2018-01-24T13:04:34

The identity is not valid.

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

Answer 2 · 2018-01-24T13:05:35

I tried this:

Have a look:

enter image source here

Answer 3 · 2018-01-24T13:08:08

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

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

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

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

#=cosx-sinx=RHS#