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(B) Given $z = \log(	an x + 	an y)$.First, we find the partial derivatives with respect to $x$ and $y$:$\frac{\partial z}{\partial x} = \frac{1}{	a

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$2$

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(B) Given $z = \log(	an x + 	an y)$.First, we find the partial derivatives with respect to $x$ and $y$:$\frac{\partial z}{\partial x} = \frac{1}{	an x + 	an y} \cdot \sec^2 x$$\frac{\partial z}{\partial y} = \frac{1}{	an x + 	an y} \cdot \sec^2 y$Now, substitute these into the expression $(\sin 2x) \frac{\partial z}{\partial x} + (\sin 2y) \frac{\partial z}{\partial y}$:$= \sin 2x \left( \frac{\sec^2 x}{	an x + 	an y} ight) + \sin 2y \left( \frac{\sec^2 y}{	an x + 	an y} ight)$$= \frac{\sin 2x \sec^2 x + \sin 2y \sec^2 y}{	an x + 	an y}$Using the identity $\sin 2	heta = 2 \sin 	heta \cos 	heta$ and $\sec^2 	heta = \frac{1}{\cos^2 	heta}$:$= \frac{(2 \sin x \cos x) \cdot \frac{1}{\cos^2 x} + (2 \sin y \cos y) \cdot \frac{1}{\cos^2 y}}{	an x + 	an y}$$= \frac{2 	an x + 2 	an y}{	an x + 	an y}$$= \frac{2(	an x + 	an y)}{	an x + 	an y} = 2$.

If $z=\log (\tan x+\tan y)$, then $(\sin 2x) \frac{\partial z}{\partial x}+(\sin 2y) \frac{\partial z}{\partial y}$ is equal to

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