If cos x - sin x = sqrt{a} sin x,then a sin x | vedclass

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(B) Given,$\cos x - \sin x = \sqrt{a} \sin x$. Rearranging the terms,we get $\cos x = \sin x(\sqrt{a} + 1)$. Multiply both sides by $(\sqrt{a} - 1)$:

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$\sqrt{a} \cos x$

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(B) Given,$\cos x - \sin x = \sqrt{a} \sin x$. Rearranging the terms,we get $\cos x = \sin x(\sqrt{a} + 1)$. Multiply both sides by $(\sqrt{a} - 1)$: $(\sqrt{a} - 1) \cos x = \sin x(\sqrt{a} + 1)(\sqrt{a} - 1)$. Using the identity $(x+y)(x-y) = x^2 - y^2$,we get: $(\sqrt{a} - 1) \cos x = \sin x(a - 1)$. $\sqrt{a} \cos x - \cos x = a \sin x - \sin x$. Rearranging to solve for $a \sin x + \cos x - \sin x$: $a \sin x + \cos x - \sin x = \sqrt{a} \cos x$.

If $\cos x - \sin x = \sqrt{a} \sin x$,then $a \sin x + \cos x - \sin x = $

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