If y = tan^{-1} left[ frac{5 cos x - 12 sin x | vedclass

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(B) Given, $y = 	an^{-1} \left[ \frac{5 \cos x - 12 \sin x}{12 \cos x + 5 \sin x} ight]$.Divide the numerator and denominator by $12 \cos x$:$y = \

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$-1$

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(B) Given, $y = 	an^{-1} \left[ \frac{5 \cos x - 12 \sin x}{12 \cos x + 5 \sin x} ight]$.Divide the numerator and denominator by $12 \cos x$:$y = 	an^{-1} \left[ \frac{\frac{5}{12} - 	an x}{1 + \frac{5}{12} 	an x} ight]$.Using the formula $	an^{-1}(A) - 	an^{-1}(B) = 	an^{-1} \left( \frac{A - B}{1 + AB} ight)$, we get:$y = 	an^{-1} \left( \frac{5}{12} ight) - 	an^{-1}(	an x)$.$y = 	an^{-1} \left( \frac{5}{12} ight) - x$.Differentiating both sides with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx} \left( 	an^{-1} \left( \frac{5}{12} ight) ight) - \frac{d}{dx}(x)$.Since $	an^{-1} \left( \frac{5}{12} ight)$ is a constant, its derivative is $0$.$\frac{dy}{dx} = 0 - 1 = -1$.

If $y = \tan^{-1} \left[ \frac{5 \cos x - 12 \sin x}{12 \cos x + 5 \sin x} \right]$, then $\frac{dy}{dx}$ is equal to

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