Question

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

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (B)

$z=\log (\tan x+\tan y)$
On differentiating partially w.r.t. $x$ and $y$, we get
$\frac{\partial z}{\partial x}=\frac{1\cdot {\sec }^{2}x}{\tan x+\tan y}$
and $\frac{\partial z}{\partial y}=\frac{{\sec }^{2}y}{\tan x+\tan y}$
Now,

sin2x∂x∂z​+sin2y∂y∂z​
$=\frac{\sin 2x{\sec }^{2}x+\sin 2y{\sec }^{2}y}{\tan x+\tan y}$
$=\frac{2\sin x\cos x\cdot \frac{1}{{\cos }^{2}x}+2\sin y\cos y\cdot \frac{1}{{\cos }^{2}y}}{\tan x+\tan y}$
$=\frac{2[\tan x+\tan y]}{\tan x+\tan y}$
$=2$