It cannot be true for  or  or any angle
which is that plus or minus a multiple of 2π or 360°. That's because
there will be a zero denominator then.  You might point that out to your
teacher that the left side is undefined in those cases.

Here's the way to find a and b for all other values of x.

 

It has to be true when x=0.

So let's plug in 0 for x:











Substitute 1 for a in

 

 

It also must be true when x = 90° or π/2

 

 

 







So a=1 and b=2

Edwin

The numerator is


    -3 + 4cos^2(x) = -3 + 4*(1-sin^2(x)) = 1 - 4sin^2(x) = (1-2sin(x))*(1+2sin(x)).



Now,   = 


             (after canceling the factor (1-2sin(x)) in the numerator and denominator)


     = 1 + 2sin(x).


Therefore,  in this identity  a= 1,  b= 2.



Surely, the identity is valid only over the domain, which is  the entire number line excluding the roots of the denominator


    1 - 2sin(x) = 0,    i.e.  except   x= arcsin(1/2) = .