Problem 19 - A Complete Resource Book in Mathematics for JEE Main

Question

The value of \(x\) for which \(\sin \left[\cot ^{1}(1+x)\right]=\cos \left(\tan ^{-1} x\right)\) is (A) \(\frac{1}{2}\) (B) 1 (C) 0 (D) \(-\frac{1}{2}\)

Step-by-Step Solution

Answer

The value of \( x \) is 0.

Step 1: Understand the problem

We are given the equation \( \sin \left[\cot ^{-1}(1+x)\right]=\cos \left(\tan ^{-1} x\right) \) and we need to find the value of \( x \) that satisfies this equation.

Step 2: Apply Property of Sine and Cosine

Recall that \( \sin(\theta) = \cos(\frac{\pi}{2} - \theta) \). Hence, equating \( \sin \left[\cot ^{-1}(1+x)\right] \) to \( \cos \left(\tan ^{-1} x\right) \), we can rewrite it as \( \cot ^{-1}(1+x) = \frac{\pi}{2} - \tan ^{-1} x \).

Step 3: Use Trigonometric Identities

Using the identities from inverse trigonometric functions, \( \cot^{-1}(y) = \tan^{-1}(\frac{1}{y}) \). Substituting, we get \( \tan^{-1}\left(\frac{1}{1+x}\right) = \frac{\pi}{2} - \tan^{-1} x \).

Step 4: Solve the Equation

From \( \tan^{-1}\left(\frac{1}{1+x}\right) = \frac{\pi}{2} - \tan^{-1} x \), add \( \tan^{-1} x \) on both sides to get \( \tan^{-1}\left(\frac{1}{1+x}\right) + \tan^{-1} x = \frac{\pi}{2} \). This means that \( \frac{1}{1+x} = \frac{1}{x} \), leading us to \( 1+x = x \), which simplifies to \( x = 0 \).

Step 5: Verify the Solution

Substitute \( x = 0 \) back into the original equations: \( \sin\left[\cot^{-1}(1)\right] = \cos\left(\tan^{-1}(0)\right) \). Evaluating both sides: \( \sin(\frac{\pi}{4}) = \cos(\frac{\pi}{2}) \), both sides are equal, confirming \( x = 0 \) is the solution.

Key Concepts

Inverse trigonometric functions

Inverse trigonometric functions allow us to determine angles when given a trigonometric ratio. They are the reverse operations of the basic trigonometric functions such as sine, cosine, and tangent. Instead of finding a trigonometric ratio from an angle, inverse functions let us find an angle from a ratio. For example:

• \( \sin^{-1}(x) \) represents the angle whose sine is \( x \).
• \( \cos^{-1}(x) \) represents the angle whose cosine is \( x \).
• \( \tan^{-1}(x) \) represents the angle whose tangent is \( x \).

The equation in our exercise involves \( \cot^{-1} \) and \( \tan^{-1} \). Because \( \cot^{-1}(y) \) and \( \tan^{-1}(x) \) are angle expressions, they follow the identity: if \( \tan^{-1}(x) + \tan^{-1}(y) = \frac{\pi}{2} \), then \( xy = 1 \). This property is crucial for solving equations where these functions appear. By understanding the nature of these inverse functions, we can tackle problems more effectively.

Trigonometric identities

Trigonometric identities are equations involving trigonometric functions that are true for every value of the involved variables. They are crucial in simplifying expressions and solving trigonometric equations. A well-known identity used in our current problem is the sine-cosine relation, where \( \sin(\theta) = \cos(\frac{\pi}{2} - \theta) \). This allows us to express sine functions in terms of cosine, facilitating easier comparison.In the context of the problem, combining this identity with the properties of inverse functions gives us:

• \( \sin[\cot^{-1}(1+x)] = \cos[\frac{\pi}{2} - \tan^{-1}(x)] \)

By understanding trigonometric identities, the equality of the sine and cosine functions can be translated into a solvable equation, highlighting the importance of these fundamental mathematical tools in problem-solving.

Solving equations

Solving trigonometric equations often involves setting equivalent expressions equal to each other to solve for the unknown variable. In our exercise, we're given \[ \tan^{-1}\left(\frac{1}{1+x}\right) + \tan^{-1}(x) = \frac{\pi}{2} \]This step leads us to use the identity \( \tan^{-1}(a) + \tan^{-1}(b) = \frac{\pi}{2} \) whenever \( ab = 1 \).Following this principle, we have:

• \( \frac{1}{1+x} \cdot x = 1 \)
• Solving \( 1 + x = x \), which simplifies to \( x = 0 \).

Ensuring every step is checked is essential. Substituting back to the original equations verifies consistency, confirming that the solution satisfies the equation. This method of solving equations requires careful manipulation of identities and properties of trigonometric functions.