If z-logtanx-tany- then sin2x- z- x-sin2y- z- y is equal to

Explanation for the correct option.Step 1. Find the value of ∂ z/∂ x and ∂ z/∂ y.Partially differentiating z=log(tanx+tany) with respect to x, we get[ ...

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Standard VII

Mathematics

Angle Sum Property

If z=logtanx+...

Question

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

$1$

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$2$

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$3$

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$4$

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Solution

The correct option is B
$2$

Explanation for the correct option.
Step 1. Find the value of $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ .
Partially differentiating $z = \log (\tan x + \tan y)$ with respect to $x$ , we get
$\begin{array}{rcl} \frac{\partial z}{\partial x} & = & \frac{\partial}{\partial x} (\log (\tan x + \tan y)) \\ = & \frac{1}{\tan x + \tan y} (\sec^{2} x + 0) \\ = & \frac{\sec^{2} x}{\tan x + \tan y} \\ = & \frac{1}{\cos^{2} x (\tan x + \tan y)} [\sec x = \frac{1}{\cos x}] \end{array}$
Partially differentiating $z = \log (\tan x + \tan y)$ with respect to $y$ , we get
$\begin{array}{rcl} \frac{\partial z}{\partial y} & = & \frac{\partial}{\partial y} (\log (\tan x + \tan y)) \\ = & \frac{1}{\tan x + \tan y} (0 + \sec^{2} y) \\ = & \frac{\sec^{2} y}{\tan x + \tan y} \\ = & \frac{1}{\cos^{2} y (\tan x + \tan y)} [\sec x = \frac{1}{\cos x}] \end{array}$
Step 2. Find the value of the expression.
In the expression $\sin 2 x \cdot (\frac{\partial z}{\partial x}) + \sin 2 y \cdot (\frac{\partial z}{\partial y})$
substituting the found values of $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ and using the formula $\sin (2 A) = \sin A \cos A$ , we get
$\begin{array}{rcl} \sin 2 x \cdot (\frac{\partial z}{\partial x}) + \sin 2 y \cdot (\frac{\partial z}{\partial y}) & = & 2 \sin x \cos x \cdot \frac{1}{\cos^{2} x (\tan x + \tan y)} + 2 \sin y \cos y \cdot \frac{1}{\cos^{2} y (\tan x + \tan y)} \\ = & \frac{2}{(\tan x + \tan y)} (\sin x \cdot \frac{1}{\cos x} + \sin y \cdot \frac{1}{\cos y}) \\ = & \frac{2}{(\tan x + \tan y)} (\frac{\sin x}{\cos x} + \frac{\sin y}{\cos y}) \\ = & \frac{2}{(\tan x + \tan y)} (\tan x + \tan y) \\ = & 2 \end{array}$
So the value of the expression $\sin 2 x \cdot (\frac{\partial z}{\partial x}) + \sin 2 y \cdot (\frac{\partial z}{\partial y})$ is $2$ .
Hence, the correct option is B.

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If z=logtanx+tany, then sin2x·∂z∂x+sin2y·∂z∂y is equal to

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