[tex]\frac{2 \sin ^{2} \alpha-1}{\sin \alpha+\cos \alpha} = sin \alpha - cos \alpha[/tex]
Solution:
Given that we have to simplify:
[tex]\frac{2 \sin ^{2} \alpha-1}{\sin \alpha+\cos \alpha}[/tex] ---- eqn 1
We know that,
[tex]sin^2 x = 1 - cos^2 x[/tex]
Substitute the above identity in eqn 1
[tex]\frac{2\left(1-\cos ^{2} \alpha\right)-1}{\sin \alpha+\cos \alpha}[/tex]
Simplify the above expression
[tex]\frac{2-2 \cos ^{2} \alpha-1}{\sin \alpha+\cos \alpha}[/tex]
[tex]\frac{1-2 \cos ^{2} \alpha}{\sin \alpha+\cos \alpha}[/tex] ------- eqn 2
By the trignometric identity,
[tex](sin x + cos x)(sin x - cos x) = 1-2cos^2 x[/tex]
Substitute the above identity in eqn 2
[tex]\frac{(\sin \alpha+\cos \alpha)(\sin \alpha-\cos \alpha)}{\sin \alpha+\cos \alpha}[/tex]
Cancel the common factors in numerator and denominator
[tex]\frac{(\sin \alpha+\cos \alpha)(\sin \alpha-\cos \alpha)}{\sin \alpha+\cos \alpha}=\sin \alpha-\cos \alpha[/tex]
Thus the simplified expression is:
[tex]\frac{2 \sin ^{2} \alpha-1}{\sin \alpha+\cos \alpha} = sin \alpha - cos \alpha[/tex]
