If y=tan ^{-1}left{frac{a cos x-b sin x}{b co | vedclass

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(C) Given $y=	an ^{-1}\left\{\frac{a \cos x-b \sin x}{b \cos x+a \sin x}ight\}$.Divide the numerator and denominator by $b \cos x$:$y=	an ^{-1}\le

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$-1$

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(C) Given $y=	an ^{-1}\left\{\frac{a \cos x-b \sin x}{b \cos x+a \sin x}ight\}$.Divide the numerator and denominator by $b \cos x$:$y=	an ^{-1}\left\{\frac{\frac{a}{b}-	an x}{1+\frac{a}{b}	an x}ight\}$.Let $\frac{a}{b}=	an \alpha$,then:$y=	an ^{-1}\left\{\frac{	an \alpha-	an x}{1+	an \alpha 	an x}ight\}$.Using the formula $	an(A-B) = \frac{	an A - 	an B}{1 + 	an A 	an B}$,we get:$y=	an ^{-1}[	an(\alpha-x)] = \alpha-x$.Differentiating with respect to $x$:$\frac{d y}{d x} = \frac{d}{d x}(\alpha-x) = 0-1 = -1$.

If $y=\tan ^{-1}\left\{\frac{a \cos x-b \sin x}{b \cos x+a \sin x}\right\}$,then find $\frac{d y}{d x}$.

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