Question

The derivative of \[{{\tan }^{-1}}\left( \dfrac{\sin x-\cos x}{\sin x+\cos x} \right)\] with respect to \[\dfrac{x}{2}\] where \[\left( x\in \left( 0,\dfrac{\pi }{2} \right) \right)\].

Answer 1

Hint: Let us assume that the value of the derivative of \[{{\tan }^{-1}}\left( \dfrac{\sin x-\cos x}{\sin x+\cos x} \right)\] with respect to \[\dfrac{x}{2}\]is equal to A. Let us assume the value of \[\dfrac{\sin x-\cos x}{\sin x+\cos x}\] is equal to B. We know that \[tan\left( A-B \right)=\dfrac{\operatorname{tanA}-\operatorname{tanB}}{1+\operatorname{tanA}\operatorname{tanB}}\]. By using this formula, we can find the value of B. Now we can find the value of the derivative of \[{{\tan }^{-1}}\left( \dfrac{\sin x-\cos x}{\sin x+\cos x} \right)\] with respect to \[\dfrac{x}{2}\]is equal to A.

Complete step-by-step answer:
From the question, it is clear that we should find the derivative of \[{{\tan }^{-1}}\left( \dfrac{\sin x-\cos x}{\sin x+\cos x} \right)\] with respect to \[\dfrac{x}{2}\]. Let us assume that the value of the derivative of \[{{\tan }^{-1}}\left( \dfrac{\sin x-\cos x}{\sin x+\cos x} \right)\] with respect to \[\dfrac{x}{2}\]is equal to A.
\[\begin{align}
  & \Rightarrow A=\dfrac{d\left( {{\tan }^{-1}}\left( \dfrac{\sin x-\cos x}{\sin x+\cos x} \right) \right)}{d\left( \dfrac{x}{2} \right)} \\
 & \Rightarrow A=\dfrac{\dfrac{d\left( {{\tan }^{-1}}\left( \dfrac{\sin x-\cos x}{\sin x+\cos x} \right) \right)}{dx}}{\dfrac{d\left( \dfrac{x}{2} \right)}{dx}}....(1) \\
\end{align}\]
Now let us simply the value of \[\dfrac{\sin x-\cos x}{\sin x+\cos x}\]. Let us assume the value of \[\dfrac{\sin x-\cos x}{\sin x+\cos x}\] is equal to B.
\[\Rightarrow B=\dfrac{\sin x-\cos x}{\sin x+\cos x}\]
Now let us take cosx as common on both numerator and denominator, then we get
\[\begin{align}
  & \Rightarrow B=\dfrac{\dfrac{\sin x}{\cos x}-1}{\dfrac{\sin x}{\cos x}+1} \\
 & \Rightarrow B=\dfrac{\tan x-1}{\tan x+1} \\
\end{align}\]
We know that the value of \[\tan \dfrac{\pi }{4}\] is equal to 1.
\[\Rightarrow B=\dfrac{\tan x-\tan \dfrac{\pi }{4}}{1+\tan x\tan \dfrac{\pi }{4}}\]
We know that \[tan\left( A-B \right)=\dfrac{\operatorname{tanA}-\operatorname{tanB}}{1+\operatorname{tanA}\operatorname{tanB}}\].
\[\Rightarrow B=\tan \left( x-\dfrac{\pi }{4} \right).....(2)\]
Let us assume the value of \[x-\dfrac{\pi }{4}\] is equal to t.
\[\Rightarrow t=x-\dfrac{\pi }{4}.....(3)\]
Now let us differentiate on both sides, then we get
\[\Rightarrow dt=dx....(4)\]
Now let us substitute equation (3) in equation (2), then we get
\[\Rightarrow B=\operatorname{tant}....(5)\]
Now let us substitute equation (4) and equation (5) in equation (1), then we get
\[\begin{align}
  & \Rightarrow A=\dfrac{\dfrac{d\left( {{\tan }^{-1}}\left( \operatorname{tant} \right) \right)}{dt}}{\dfrac{d\left( \dfrac{x}{2} \right)}{dx}} \\
 & \Rightarrow A=\dfrac{\dfrac{dt}{dt}}{\dfrac{d\left( \dfrac{x}{2} \right)}{dx}} \\
 & \Rightarrow A=\dfrac{1}{\left( \dfrac{1}{2} \right)} \\
 & \Rightarrow A=2.......(6) \\
\end{align}\]
From equation (6), it is clear that the value of A is equal to 2.
So, it is clear that the derivative of \[{{\tan }^{-1}}\left( \dfrac{\sin x-\cos x}{\sin x+\cos x} \right)\] with respect to \[\dfrac{x}{2}\] where \[\left( x\in \left( 0,\dfrac{\pi }{2} \right) \right)\] is equal to 2.

So, the correct answer is “Option D”.

Note: Students may have a misconception that \[tan\left( A+B \right)=\dfrac{\operatorname{tanA}-\operatorname{tanB}}{1+\operatorname{tanA}\operatorname{tanB}}\]. But we know that \[tan\left( A-B \right)=\dfrac{\operatorname{tanA}-\operatorname{tanB}}{1+\operatorname{tanA}\operatorname{tanB}}\]. So, this misconception should be avoided by students. If this misconception is followed, then it will interrupt the solution and will affect the final answer. So, this misconception should be avoided.