If frac{sin^2 x + 1}{2sin^2 x - 5sin x + 3} = | vedclass

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(D) Let $u = \sin x$. The expression is $\frac{u^2 + 1}{2u^2 - 5u + 3} = \frac{u^2 + 1}{(2u - 3)(u - 1)}$.Using partial fractions,we write $\frac{u^2

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$A$ and $B$ both

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(D) Let $u = \sin x$. The expression is $\frac{u^2 + 1}{2u^2 - 5u + 3} = \frac{u^2 + 1}{(2u - 3)(u - 1)}$.Using partial fractions,we write $\frac{u^2 + 1}{(2u - 3)(u - 1)} = C + \frac{A}{2u - 3} + \frac{B}{u - 1}$.Since the degree of the numerator equals the degree of the denominator,$C$ is the ratio of the leading coefficients: $C = \frac{1}{2}$.Now,$\frac{u^2 + 1}{(2u - 3)(u - 1)} = \frac{1}{2} + \frac{A}{2u - 3} + \frac{B}{u - 1}$.$\frac{u^2 + 1 - \frac{1}{2}(2u^2 - 5u + 3)}{(2u - 3)(u - 1)} = \frac{\frac{5}{2}u - \frac{1}{2}}{(2u - 3)(u - 1)} = \frac{A}{2u - 3} + \frac{B}{u - 1}$.Using the cover-up method:For $B$: Multiply by $(u - 1)$ and set $u = 1$: $B = \frac{\frac{5}{2}(1) - \frac{1}{2}}{2(1) - 3} = \frac{2}{-1} = -2$.For $A$: Multiply by $(2u - 3)$ and set $u = \frac{3}{2}$: $A = \frac{\frac{5}{2}(\frac{3}{2}) - \frac{1}{2}}{\frac{3}{2} - 1} = \frac{\frac{15}{4} - \frac{2}{4}}{\frac{1}{2}} = \frac{13}{4} 	imes 2 = \frac{13}{2}$.Thus,$A = \frac{13}{2}$,$B = -2$,$C = \frac{1}{2}$.$A + B + C = \frac{13}{2} - 2 + \frac{1}{2} = 7 - 2 = 5$.Both $(A)$ and $(B)$ are correct.

If $\frac{\sin^2 x + 1}{2\sin^2 x - 5\sin x + 3} = \frac{A}{2\sin x - 3} + \frac{B}{\sin x - 1} + C$,then:

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