Question

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

• 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,

$=\frac{\sin 2 x \sec ^{2} x+\sin 2 y \sec ^{2} y}{\tan x+\tan y}$
$=\frac{2 \sin x \cos x \cdot \frac{1}{\cos ^{2} x}+2 \sin y \cos y \cdot \frac{1}{\cos ^{2} y}}{\tan x+\tan y}$
$=\frac{2[\tan x+\tan y]}{\tan x+\tan y}$
$=2$