If y = tan^{-1}left( frac{acos x - bsin x}{bc | vedclass

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(B) 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}\l

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$-1$

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(B) 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 $\frac{a}{b} = 	an 	heta$,then $y = 	an^{-1}\left( \frac{	an 	heta - 	an x}{1 + 	an 	heta 	an x} ight)$.Using the formula $	an(A - B) = \frac{	an A - 	an B}{1 + 	an A 	an B}$,we get:$y = 	an^{-1}(	an(	heta - x)) = 	heta - x$.Since $	heta = 	an^{-1}(\frac{a}{b})$ is a constant,differentiating with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx}(	an^{-1}(\frac{a}{b}) - x) = 0 - 1 = -1$.

If $y = \tan^{-1}\left( \frac{a\cos x - b\sin x}{b\cos x + a\sin x} \right)$,then $\frac{dy}{dx} = $

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