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

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.

& \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{aligned}$$ Now let us simply the value of $$\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{aligned} & \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{aligned}$$ We know that the value of $$\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{aligned} & \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{aligned}$$ From equation (6), it is clear that the value of A is equal to 2. So, it is clear that the derivative of $${{\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.