Question

If \[z = \log \left( {\tan x + \tan y} \right)\], then \[\sin 2x(\dfrac{{\partial z}}{{\partial x}}) + \sin 2y(\dfrac{{\partial z}}{{\partial y}})\] is equal to
A. \[1\]
B. \[2\]
C. \[3\]
D. \[4\]

Answer 1

Hint : We are given a function \[z = \log \left( {\tan x + \tan y} \right)\] . We have to find the value of \[\sin 2x(\dfrac{{\partial z}}{{\partial x}}) + \sin 2y(\dfrac{{\partial z}}{{\partial y}})\]. We will first consider that \[\log = \ln \], then we will be taking antilog on both sides of the equation of the given function. Then we will differentiate the obtained antilog equation using partial differentiation. Then find the value of \[\sin 2x(\dfrac{{\partial z}}{{\partial x}})\] and \[\sin 2y(\dfrac{{\partial z}}{{\partial y}})\], then add them to find the required value.

Formula used : Following formulae would be useful for solving this question
Antilog of \[z = \ln x\] is \[x = {e^z}\]
\[
  \sin 2x = 2\sin x\cos x \\
  \dfrac{\partial }{{\partial x}}(\tan x) = {\sec ^2}x \\
 \]
Complete Step-by-Step Solution :
We are given a function \[z = \log \left( {\tan x + \tan y} \right)\]
Assume, \[\log = \ln \],
Given function becomes \[z = \ln \left( {\tan x + \tan y} \right)\]
On taking the antilog on both sides of the equation, we get
\[{e^z} = \tan x + \tan y\] …… (1)
Now, on partially differentiating equation (1) w.r.t \[x\], we get
\[
  {e^z}\dfrac{{\partial z}}{{\partial x}} = {\sec ^2}x + 0 \\
  \dfrac{{\partial z}}{{\partial x}} = \dfrac{{{{\sec }^2}x}}{{{e^z}}} \\
 \]
On multiplying the above the equation with \[\sin 2x\] and substituting \[\sin 2x = 2\sin x\cos x\] in the RHS of the obtained equation, we get
\[\sin 2x\dfrac{{\partial z}}{{\partial x}} = \sin 2x\dfrac{{{{\sec }^2}x}}{{{e^z}}}\]
On substituting \[\sin 2x = 2\sin x\cos x\] in the RHS, we get
\[
  \sin 2x\dfrac{{\partial z}}{{\partial x}} = 2\sin x\cos x \times \dfrac{{{{\sec }^2}x}}{{{e^z}}} \\
   = \dfrac{{2\sin x\cos x}}{{{{\cos }^2}x \times {e^z}}} \\
 \]
On further solving, we get
\[\sin 2x\dfrac{{\partial z}}{{\partial x}} = 2\tan x \times {e^{ - z}}\] …… (2)
Similarly,
\[\sin 2x\dfrac{{\partial z}}{{\partial y}} = 2\tan y \times {e^{ - z}}\] …… (3)
Now, on summing the equations (2) and (3), we get
\[
  \sin 2x(\dfrac{{\partial z}}{{\partial x}}) + \sin 2y(\dfrac{{\partial z}}{{\partial y}}) = 2\tan x \times {e^{ - z}} + 2\tan y \times {e^{ - z}} \\
   = 2{e^{ - z}}(\tan x + \tan y) \\
 \]
On putting the value from equation (1) in the above equation, we get

\[
  \sin 2x(\dfrac{{\partial z}}{{\partial x}}) + \sin 2y(\dfrac{{\partial z}}{{\partial y}}) = 2{e^{ - z}}(\tan x + \tan y) \\
   = 2{e^{ - z}}{e^z} \\
   = 2{e^{ - z + z}} \\
   = 2{e^0} \\
 \]
On further solving, we get
\[\sin 2x(\dfrac{{\partial z}}{{\partial x}}) + \sin 2y(\dfrac{{\partial z}}{{\partial y}}) = 2\]

Thus the correct option is B.

Note : In this question, students may make mistake while differentiating \[\tan x\]. Instead of \[{\sec ^2}x\] they may write \[\sec x\] , which is wrong. Differentiation of \[\tan x\] is \[{\sec ^2}x\]. Also, here, students may make mistake after multiplying \[\sin 2x\]and substituting it as \[\sin 2x = 2\sin x\cos x\]. After performing this step, they should substitute \[{\sec ^2}x = \dfrac{1}{{{{\cos }^2}x}}\] to simplify further and get the required value.