frac{1-cos(2x)+sin(x)}{sin(2x)+cos(x)} = | vedclass

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(C) Given expression: $\frac{1-\cos(2x)+\sin(x)}{\sin(2x)+\cos(x)}$Using the identity $\cos(2x) = 1 - 2\sin^2(x)$,the numerator becomes: $1 - (1 - 2\s

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$	an(x)$

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(C) Given expression: $\frac{1-\cos(2x)+\sin(x)}{\sin(2x)+\cos(x)}$Using the identity $\cos(2x) = 1 - 2\sin^2(x)$,the numerator becomes: $1 - (1 - 2\sin^2(x)) + \sin(x) = 2\sin^2(x) + \sin(x) = \sin(x)(2\sin(x) + 1)$Using the identity $\sin(2x) = 2\sin(x)\cos(x)$,the denominator becomes: $2\sin(x)\cos(x) + \cos(x) = \cos(x)(2\sin(x) + 1)$Substituting these back into the expression: $\frac{\sin(x)(2\sin(x) + 1)}{\cos(x)(2\sin(x) + 1)} = \frac{\sin(x)}{\cos(x)} = 	an(x)$

$\frac{1-\cos(2x)+\sin(x)}{\sin(2x)+\cos(x)} = $

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