If y = tan^{-1} left( frac{1 - cos 3x}{sin 3x | vedclass

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(C) Given,$y = 	an^{-1} \left( \frac{1 - \cos 3x}{\sin 3x} ight)$.Using trigonometric identities $1 - \cos 	heta = 2 \sin^2 \left( \frac{	heta}{2

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$\frac{3}{2}$

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(C) Given,$y = 	an^{-1} \left( \frac{1 - \cos 3x}{\sin 3x} ight)$.Using trigonometric identities $1 - \cos 	heta = 2 \sin^2 \left( \frac{	heta}{2} ight)$ and $\sin 	heta = 2 \sin \left( \frac{	heta}{2} ight) \cos \left( \frac{	heta}{2} ight)$,we have:$y = 	an^{-1} \left( \frac{2 \sin^2 \left( \frac{3x}{2} ight)}{2 \sin \left( \frac{3x}{2} ight) \cos \left( \frac{3x}{2} ight)} ight)$$y = 	an^{-1} \left( \frac{\sin \left( \frac{3x}{2} ight)}{\cos \left( \frac{3x}{2} ight)} ight)$$y = 	an^{-1} \left( 	an \left( \frac{3x}{2} ight) ight)$$y = \frac{3x}{2}$Differentiating with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx} \left( \frac{3x}{2} ight) = \frac{3}{2}$.

If $y = \tan^{-1} \left( \frac{1 - \cos 3x}{\sin 3x} \right)$,then $\frac{dy}{dx} = \ldots$

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