Problem 72 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The derivative of \(\tan ^{-1}\left(\frac{\sin x-\cos x}{\sin x+\cos x}\right)\), with respect to \(\frac{x}{2}\), where \(\left(x \in\left(0, \frac{\pi}{2}\right)\right)\) is: \(\quad\) (a) 1 (b) \(\frac{2}{3}\) (c) \(\frac{1}{2}\) (d) 2

Step-by-Step Solution

Answer

The derivative is 1, option (a).

Step 1: Simplify the Inner Function

First, let's simplify the expression \( \frac{\sin x - \cos x}{\sin x + \cos x} \). This can be rewritten using the identity \( \sin x = \frac{2 \tan \left( \frac{x}{2} \right)}{1 + \tan^2 \left( \frac{x}{2} \right)} \) and \( \cos x = \frac{1 - \tan^2 \left( \frac{x}{2} \right)}{1 + \tan^2 \left( \frac{x}{2} \right)} \). Applying these identities, the expression simplifies to \( \tan\left(\frac{x}{2}\right) \). Thus, \( \tan^{-1}\left(\frac{\sin x - \cos x}{\sin x + \cos x}\right) = \tan^{-1}\left(\tan\left(\frac{x}{2}\right)\right) = \frac{x}{2} \).

Step 2: Calculate the Derivative

We need to differentiate the function \( \frac{x}{2} \) with respect to \( \frac{x}{2} \). Since this is a linear function, the derivative is computed easily as \( \frac{d}{d(\frac{x}{2})}\left(\frac{x}{2}\right) = 1 \).

Step 3: Verify the Solution

The solution is verified by noting that differentiating a function \( y = ax \) with respect to \( x \) gives the constant \( a \). Here, \( a = 1 \) because \( y = \frac{x}{2} \) and we are differentiating with respect to \( \frac{x}{2} \). The verification confirms that the derivative of \( \frac{x}{2} \) with respect to itself is indeed \( 1 \).

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are essential to calculus because they allow us to determine angles given a ratio of sides in right triangles. They include functions like \(\tan^{-1}(x)\), \(\sin^{-1}(x)\), and \(\cos^{-1}(x)\). The function \(\tan^{-1}(x)\) is the inverse of the trigonometric function tangent, which means for an angle \( y \), \( \tan(y) = x \) is equivalent to \( y = \tan^{-1}(x)\).
This concept is used in the original exercise to convert a complex fraction involving \( \sin x \) and \( \cos x \) into a simpler form involving \( x \) directly. Specifically, we use trigonometric identities to simplify an expression that can then be expressed directly as the angle \( \frac{x}{2} \), thanks to the inverse tangent function.
By applying \(\tan^{-1}\) to the expression, we make it possible to more easily differentiate later in the problem, showcasing the inverse part's critical role in managing complex trigonometric forms.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are always true for all values of the occurring variables where both sides of the equality are defined. They are handy tools in simplifying expressions and solving equations involving trigonometric functions.
In our exercise, two critical identities are used:

• \( \sin x = \frac{2 \tan \left( \frac{x}{2} \right)}{1 + \tan^2 \left( \frac{x}{2} \right)} \)
• \( \cos x = \frac{1 - \tan^2 \left( \frac{x}{2} \right)}{1 + \tan^2 \left( \frac{x}{2} \right)} \)

These identities express sine and cosine in terms of the tangent of half the angle \( \frac{x}{2} \). Such representations are vital because they simplify the manipulation of expressions, leading to a more straightforward calculation when finding derivatives.
Using these identities, our initial complex expression \( \frac{\sin x - \cos x}{\sin x + \cos x} \) transforms into \( \tan \left( \frac{x}{2} \right) \). This step significantly simplifies the differentiation that follows.

Calculus Differentiation

Differentiation in calculus refers to finding the rate at which a function changes at any given point. It provides the slope of the tangent line to the curve of the function at that point. In the context of the exercise, once we have the simplified function \( \tan^{-1}\left(\frac{\sin x - \cos x}{\sin x + \cos x}\right) = \frac{x}{2} \), we need to find its derivative with respect to the variable \( \frac{x}{2} \).
Since \( \frac{x}{2} \) is a linear function, its differentiation is straightforward. The derivative process reveals how the function scales with its input, which is crucial in calculus operations. When differentiating \( y = ax \) with respect to \( x \), the derivative is the constant \( a \), indicating how much \( y \) changes per unit change in \( x \).
In our solution, by differentiating \( \frac{x}{2} \) with respect to itself, we confirm that the derivative is indeed 1. This result aligns with the key concept that the rate of change of a linear function where the output is identical to the input variable is always 1, providing a solid foundation in understanding linear relationships in calculus.