If y = cot^{-1} left( frac{1 + sin 5x}{cos 5x | vedclass

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(D) Given $y = \cot^{-1} \left( \frac{1 + \sin 5x}{\cos 5x} ight)$.Using trigonometric identities $1 + \sin 	heta = \left( \cos \frac{	heta}{2} +

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$-\frac{5}{2}$

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(D) Given $y = \cot^{-1} \left( \frac{1 + \sin 5x}{\cos 5x} ight)$.Using trigonometric identities $1 + \sin 	heta = \left( \cos \frac{	heta}{2} + \sin \frac{	heta}{2} ight)^2$ and $\cos 	heta = \cos^2 \frac{	heta}{2} - \sin^2 \frac{	heta}{2}$, we have:$y = \cot^{-1} \left( \frac{(\cos \frac{5x}{2} + \sin \frac{5x}{2})^2}{(\cos \frac{5x}{2} - \sin \frac{5x}{2})(\cos \frac{5x}{2} + \sin \frac{5x}{2})} ight)$$y = \cot^{-1} \left( \frac{\cos \frac{5x}{2} + \sin \frac{5x}{2}}{\cos \frac{5x}{2} - \sin \frac{5x}{2}} ight)$Divide numerator and denominator by $\cos \frac{5x}{2}$:$y = \cot^{-1} \left( \frac{1 + 	an \frac{5x}{2}}{1 - 	an \frac{5x}{2}} ight) = \cot^{-1} \left( 	an \left( \frac{\pi}{4} + \frac{5x}{2} ight) ight)$Since $\cot^{-1}(	an 	heta) = \cot^{-1}(\cot(\frac{\pi}{2} - 	heta))$, we get:$y = \frac{\pi}{2} - (\frac{\pi}{4} + \frac{5x}{2}) = \frac{\pi}{4} - \frac{5x}{2}$Therefore, $\frac{dy}{dx} = \frac{d}{dx} (\frac{\pi}{4} - \frac{5x}{2}) = -\frac{5}{2}$.

If $y = \cot^{-1} \left( \frac{1 + \sin 5x}{\cos 5x} \right)$, then the value of $\frac{dy}{dx}$ is

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