Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
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
MathematicsDifferentiationAP EAMCETAP EAMCET 2007
Options:
  • A $1$
  • B $2$
  • C $3$
  • D $4$
Solution:
1129 Upvotes Verified Answer
The correct answer is: $2$
$z=\log (\tan x+\tan y)$
On differentiating partially w.r.t. $x$ and $y$, we get
and $\begin{aligned} \frac{\partial z}{\partial x} & =\frac{1 \cdot \sec ^2 x}{\tan x+\tan y} \\ \text { and } \quad \frac{\partial z}{\partial y} & =\frac{\sec ^2 y}{\tan x+\tan y}\end{aligned}$
Now,
$\begin{aligned} & \sin 2 x \frac{\partial z}{\partial x}+\sin 2 y \frac{\partial z}{\partial y} \\ & =\frac{\sin 2 x \sec ^2 x+\sin 2 y \sec ^2 y}{\tan x+\tan y} \\ & =\frac{2 \sin x \cos x \cdot \frac{1}{\cos ^2 x}+2 \sin y \cos y \cdot \frac{1}{\cos ^2 y}}{\tan x+\tan y} \\ & =\frac{2[\tan x+\tan y]}{\tan x+\tan y} \\ & =2\end{aligned}$

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.