Explanation:
Use the Pythagorean identity, cancel common factors, and divide numerator and denominator by cos(x). Equivalently, multiply numerator and denominator by sec(x).
[tex]\dfrac{\sin^2{x}+2\cos{x}-1}{\sin^2{x}+3\cos{x}-3}=\dfrac{1-\cos^2{x}+2\cos{x}-1}{1-\cos^2{x}+3\cos{x}-3} \qquad\text{replace $\sin^2$ with $1-\cos^2$}\\\\=\dfrac{\cos{x}(2-\cos{x})}{(\cos{x}-1)(2-\cos{x})}=\dfrac{\cos{x}}{\cos{x}-1}=\dfrac{1}{\dfrac{\cos{x}}{\cos{x}}-\dfrac{1}{\cos{x}}}=\dfrac{1}{1-\sec{x}}[/tex]
