Problem 38 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(A=\cot ^{-1} \sqrt{\tan \theta}-\tan ^{-1} \sqrt{\tan \theta}\), then \(\tan \left(\frac{\pi}{4}-\frac{A}{2}\right)\) is equal to (A) \(\sqrt{\cot \theta}\) (B) \(\tan \theta\) (C) \(\sqrt{\tan \theta}\) (D) none of these

Step-by-Step Solution

Answer

The answer is \(\tan \theta\), which coincides with option B.

Step 1: Express A in terms of a single trigonometric function

We start by expressing \( A = \cot^{-1}(\sqrt{\tan \theta}) - \tan^{-1}(\sqrt{\tan \theta}) \). Recognize that if \( \alpha = \tan^{-1}(\sqrt{\tan \theta}) \), then \( A = \cot^{-1}(\sqrt{\tan \theta}) - \alpha \). From the property of inverse trigonometric functions, we have \( \tan(\alpha) = \sqrt{\tan \theta} \) and thus, \( \alpha = \tan^{-1}(\sqrt{\tan \theta}) \). The cotangent inverse gives us \( \cot(A + \alpha) = \sqrt{\tan \theta} \).

Step 2: Find relationship between cotangent of sum and tangent

Using the relation \( \cot(B) = \tan(\pi/2 - B) \), we translate \( \cot(A + \alpha) = \sqrt{\tan \theta} \) to \( \tan(\pi/2 - (A + \alpha)) = \sqrt{\tan \theta} \). This means that \( \pi/2 - (A + \alpha) = \tan^{-1}(\sqrt{\tan \theta}) \).

Step 3: Simplify \( \frac{A}{2} \) and substitute

First, calculate \( A \) from the expression derived in previous steps: \( \pi/2 - (A + \alpha) = \tan^{-1}(\sqrt{\tan \theta}) \) implies \( A \) is zero because \( \tan^{-1}(\sqrt{\tan \theta}) + \tan^{-1}(\sqrt{\tan \theta}) = \pi/2 \) under the condition that \( 2\tan^{-1}(\sqrt{\tan \theta}) = \pi/2 \). Justify this under typical domain and range considerations for inverse trigonometric functions.

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Step 5: Simplify using half-angle formula

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Key Concepts

Trigonometric Identities

Understanding trigonometric identities is crucial when solving problems involving inverse trigonometric functions. These identities provide relationships between different trigonometric functions, helping to simplify complex expressions.

• The Pythagorean identities, such as \( \sin^2 \theta + \cos^2 \theta = 1 \), are foundational, connecting sines and cosines.
• Reciprocal identities, like \( \cot \theta = \frac{1}{\tan \theta} \), show relationships between trigonometric functions and their reciprocals.
• Inverse trigonometric functions provide identities like \( \tan^{-1}(x) + \cot^{-1}(x) = \frac{\pi}{2} \). This identity is particularly useful because it simplifies expressions by combining inverse tangents and inverse cotangents.

Using these identities, particularly with inverse functions, allows us to rewrite and simplify trigonometric equations more easily. This skill is pivotal for solving the original exercise, where identities help transform the expression into a simpler form.

Trigonometric Equations

Trigonometric equations involve solving for an angle that satisfies a given trigonometric expression. It's akin to a regular equation but centered around angles and their trigonometric counterparts.

• Solving such equations often requires isolating a trigonometric function on one side using algebraic manipulations.
• Inverse functions play an integral role, translating specific values back into angles, like converting \( \tan^{-1}(x) \) back to \( \theta \).
• Understanding the periodic nature of trigonometric functions helps recognize that there can be multiple solutions within a given interval.

In the given exercise, we use these equations to determine the angle \( A \) by ensuring the terms involving \( \tan \) and \( \cot \) align with known values, reducing the complete trigonometric expression to evaluate \( \tan \left( \frac{\pi}{4} - \frac{A}{2} \right) \). This requires a fluency with both understanding inverse trigonometric functions and manipulating trigonometric identities.

Half-Angle Formulas

Half-angle formulas are specific trigonometric identities that express trigonometric functions of half angles in terms of the function of the full angle. They are derived from the double angle formulas and are very useful in simplifying problems where angles are halved.

• The half-angle formula for tangent is: \( \tan\left(\frac{x}{2}\right) = \frac{1 - \cos x}{\sin x} \)
• These formulas allow us to break down angles like \( \frac{\pi}{4} - \frac{A}{2} \) into more manageable terms.
• The use of half-angle formulas can significantly simplify problems that involve inverse trigonometric functions and lead to exact answers.

In the context of the problem, simplifying \( \tan\left(\frac{\pi}{4} - \frac{A}{2}\right) \) using half-angle identities clarifies what happens when relations between tangent and cotangent are established, offering a route to the solution by breaking down complex trigonometric terms into half angles. This particular operation transformed our exercise into a more digestible form, ultimately leading to the solution.