Question 1177724
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<pre>

The numerator is


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



Now,  {{{(-3+4cos^2(x))/(1-2sin(x))}}} = 


             (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) = {{{pi/6 + 2k*pi}}}.         <U>ANSWER</U>
</pre>


Solved, answered and explained.   &nbsp;&nbsp;&nbsp;&nbsp;And completed.



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This problem was posted to the forum a week or two ago, and I solved it under this link


<A HREF=https://www.algebra.com/algebra/homework/Trigonometry-basics/Trigonometry-basics.faq.question.1177378.html>https://www.algebra.com/algebra/homework/Trigonometry-basics/Trigonometry-basics.faq.question.1177378.html</A>


https://www.algebra.com/algebra/homework/Trigonometry-basics/Trigonometry-basics.faq.question.1177378.html



Here I simply pasted and copied that my solution for your convenience.



E N J O Y (!)