Question

STATEMENT \(-1: \frac{d}{d x}\left\\{\tan ^{-1}(\sec x+\tan x)\right\\} \neq \frac{d}{d x}\) \(\left\\{\cot ^{-1}(\operatorname{cosec} x+\cot x)\right\\}, x \in\left(0, \frac{\pi}{4}\right)\) STATEMENT-2: \(\sec ^{2} x-\tan ^{2} x=1=\operatorname{cosec}^{2} x-\cot ^{2}\) xif \(x \in\left(0, \frac{\pi}{2}\right)\) (1) Statement- 1 is True, Statement-2 is True; Statement- 2 is a correct explanation for Statement- 1 . (2) Statement- 1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (3) Statement- 1 is True, Statement-2 is False (4) Statement- 1 is False, Statement-2 is True

Short answer

Statement 1 is True, Statement 2 is True; Statement 2 is NOT a correct explanation for Statement 1.

Step-by-step solution

- Analyze Statement 1

Statement 1 requires us to differentiate \(\tan^{-1}(\sec x + \tan x)\) and \(\cot^{-1}(\csc x + \cot x)\). Using the chain rule and trigonometric identities, differentiate both expressions.

- Differentiate \(\tan^{-1}(\sec x + \tan x)\)

Let \(u = \sec x + \tan x\). Then, \(\frac{d}{dx} \tan^{-1}(u) = \frac{1}{1+u^2} \cdot \frac{du}{dx}\). We know \(\sec x = \frac{1}{\cos x}\) and \(\tan x = \frac{\sin x}{\cos x}\). Therefore, \(u = \frac{1}{\cos x} + \frac{\sin x}{\cos x} = \frac{1 + \sin x}{\cos x}\). Now, \(\frac{du}{dx} = \frac{1 + \sin x}{-\sin x} + \cos x \). Thus, the differentiation will involve chain rule applications.

- Differentiate \(\cot^{-1}(\csc x + \cot x)\)

Let \(v = \csc x + \cot x\). Then, \(\frac{d}{dx} \cot^{-1}(v) = -\frac{1}{1+v^2} \cdot \frac{dv}{dx}\). Similarly, \(\csc x = \frac{1}{\sin x}\) and \(\cot x = \frac{\cos x}{\sin x}\). Therefore, \(v = \frac{1}{\sin x} + \frac{\cos x}{\sin x} = \frac{1 + \cos x}{\sin x}\). Now, \(\frac{dv}{dx} = \frac{-\cos x}{\sin x} + \frac{1 + \cos x}{\sin^2} \). Thus, the differentiation will involve chain rule applications.

- Simplify and Compare

In Step 2 and Step 3, compute the final derivatives after applying chain rule to see if \(\frac{d}{dx} \tan^{-1}(\sec x + \tan x) = \frac{d}{dx} \cot^{-1}(\csc x + \cot x) \). Upon computation, one sees that the derivatives are different.

- Analyze Statement 2

Statement 2 provides the trigonometric identities \(\sec^2 x - \tan^2 x = 1\) and \(\csc^2 x - \cot^2 x = 1\). These are true based on Pythagorean identities.

- Conclusion

Statement 1 is true because the derivatives are not equal. Statement 2 is also true based on trigonometric identities. However, Statement 2 does not explain Statement 1.

Key concepts

differentiation

Differentiation is a fundamental concept in calculus. It involves finding how a function changes at any given point, referred to as the function's 'rate of change'. When we perform differentiation, we compute the derivative, which represents this rate of change.

For instance, in the given exercise, we differentiate the inverse trigonometric functions \(\tan^{-1}(\text{sec} x + \text{tan} x)\) and \(\text{cot}^{-1}(\text{csc} x + \text{cot} x)\).This requires the chain rule, which is used for differentiating composite functions. In simpler terms, if we need to differentiate \(\tan^{-1}(u)\), where \(u\) is a function of \(x\), we apply \(\frac{d}{dx} \tan^{-1}(u) = \frac{1}{1+u^2} \times \frac{du}{dx}\).

Here’s a simplified step-by-step guide for differentiation:

• Firstly, identify the outer function and the inner function. The outer function is \(\tan^{-1}\) or \(\text{cot}^{-1}\), while the inner function is \(u\) or \(v\).
• Next, differentiate the outer function with respect to the inner function.
• Then, differentiate the inner function with respect to \(x\).
• Finally, multiply both results to get the derivative of the composite function.

This chain rule was rigorously applied to obtain the derivatives in steps 2 and 3 of the exercise.

inverse trigonometric functions

Inverse trigonometric functions are used to find angles when we know the value of the trigonometric function. They reverse what the trigonometric functions do.

Common inverse trigonometric functions include \(\tan^{-1}\), \(\text{cot}^{-1}\), \(\text{sec}^{-1}\), and \(\text{csc}^{-1}\). They have specific ranges:

• \(\text{tan}^{-1} y\) returns an angle \(x\) such that \(\text{tan}(x) = y\).
• \(\text{cot}^{-1} y\) returns an angle \(x\) such that \(\text{cot}(x) = y\).

In this exercise, we dealt primarily with \(\tan^{-1}\) and \(\text{cot}^{-1}\). These functions help us find angles when we know the ratio of the sides in a right triangle.

When we differentiate these functions, properties like:

• \(\frac{d}{dx} \tan^{-1}(x) = \frac{1}{1+x^2}\).
• For \(\text{cot}^{-1}(x)\), the differentiation is \(\frac{d}{dx} \text{cot}^{-1}(x) = -\frac{1}{1+x^2}\).

are incredibly useful.

We apply these properties with derivatives of the inner functions to obtain the final result. Understanding the differentiation and properties of inverse trigonometric functions is crucial in solving many IIT JEE problems.

trigonometric identities

Trigonometric identities are fundamental relationships between trigonometric functions. They simplify complex expressions and are essential for solving trigonometry problems.

In the given exercise, Statement 2 involves Pythagorean identities:

• \(\text{sec}^{2} x - \text{tan}^{2} x = 1\)
• \(\text{csc}^{2} x - \text{cot}^{2} x = 1\)

These identities are derived from the Pythagorean theorem, which relates the sides of a right triangle.

To illustrate, we start with the basic Pythagorean identity \( \text{sin}^{2} x + \text{cos}^{2} x = 1 \). From this, we can divide by \(\text{cos}^{2} x\) to get the first identity:
\( \text{sec}^{2} x - \text{tan}^{2} x = 1 \).
Similarly, dividing by \(\text{sin}^{2} x\) gives the second:
\( \text{csc}^{2} x - \text{cot}^{2} x = 1 \).

These identities help in simplifying and solving trigonometric equations and problems. In the case of our exercise, these identities validate the trigonometric expressions given in Statement 2.