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

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$2$

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

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