Problem 13 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The value of \(\sin ^{-1}\left(\frac{12}{13}\right)-\sin ^{-1}\left(\frac{3}{5}\right)\) is equal to : (a) \(\pi-\sin ^{-1}\left(\frac{63}{65}\right)\) (b) \(\frac{\pi}{2}-\sin ^{-1}\left(\frac{56}{65}\right)\) (c) \(\frac{\pi}{2}-\cos ^{-1}\left(\frac{9}{65}\right)\) (d) \(\pi-\cos ^{-1}\left(\frac{33}{65}\right)\)

Step-by-Step Solution

Answer

The correct option is (d) \(\pi - \cos^{-1}\left(\frac{33}{65}\right)\).

Step 1: Understanding Inverse Sine and Sine Addition Formulas

The problem involves inverse sine (\( ext{sin}^{-1}\)) functions and their differences. We aim to simplify the expression using properties of trigonometric identities, specifically the formula for the difference of inverse sine:\[\sin^{-1}x - \sin^{-1}y = \sin^{-1}\left(x\sqrt{1-y^2} - y\sqrt{1-x^2}\right)\]Understanding this core formula is essential for solving the problem.

Step 2: Simplifying using the Formula

We'll substitute the given values into the formula: \(x = \frac{12}{13}\) and \(y = \frac{3}{5}\). First, compute the necessary components:\[\sqrt{1-x^2} = \sqrt{1-\left(\frac{12}{13}\right)^2} = \sqrt{\frac{25}{169}} = \frac{5}{13}\]\[\sqrt{1-y^2} = \sqrt{1-\left(\frac{3}{5}\right)^2} = \sqrt{\frac{16}{25}} = \frac{4}{5}\]Now substitute these into the difference formula:\[\sin^{-1}\left(x\sqrt{1-y^2} - y\sqrt{1-x^2}\right) = \sin^{-1}\left(\frac{12}{13} \cdot \frac{4}{5} - \frac{3}{5} \cdot \frac{5}{13}\right)\]

Step 3: Calculate the Simplified Expression

Calculate the expression inside \( ext{sin}^-1\):\[\frac{12}{13} \cdot \frac{4}{5} = \frac{48}{65}\]\[\frac{3}{5} \cdot \frac{5}{13} = \frac{15}{65}\]Now, evaluate and simplify:\[\frac{48}{65} - \frac{15}{65} = \frac{33}{65}\]So, we have:\[\sin^{-1}\left(\frac{33}{65}\right)\]

Step 4: Compare with Answer Options

Now that we have \(\sin^{-1}\left(\frac{33}{65}\right)\), compare it with the given answer options. It matches the choice \(\pi - \cos^{-1}\left(\frac{33}{65}\right)\), since for any angle \(\theta\):\[\cos(\theta) = \sin(\pi/2 - \theta)\]Thus, \(\pi - \cos^{-1}\left(\frac{33}{65}\right) = \sin^{-1}\left(\frac{33}{65}\right)\).This matches option (d).

Key Concepts

Sine Addition Formula

The sine addition formula is pivotal in solving complex trigonometric problems, especially those involving inverse sine functions like \(\sin^{-1}\). When dealing with inverse sine, or arcsine, functions, understanding how to combine or separate them efficiently becomes crucial. The arcsine of a difference can be rewritten using the formula:\[\sin^{-1}(x) - \sin^{-1}(y) = \sin^{-1}\left(x\sqrt{1-y^2} - y\sqrt{1-x^2}\right)\]This formula essentially allows you to calculate the difference between two inverse sine values by transforming them into a single expressions involving square roots and simpler trigonometric components.

• Useful for breaking down complex expressions
• Simplifies solving processes

Recognizing and applying this formula appropriately turns a seemingly complicated expression into something more manageable.

Trigonometric Identities

Trigonometric identities are foundational tools in trigonometry that enable the transformation and simplification of expressions. These identities are equations that hold for any angles, making them very versatile.
In this exercise, we utilize a special identity involving arcsine to bridge the gap between sine and cosine. Generally, trigonometric identities help relate different trigonometric functions and angles:

• They offer methods to convert between sine, cosine, and other functions.
• Useful in rewriting and simplifying expressions for easier calculations.

Grasping these identities, such as the Pythagorean identity or angle difference identities, allows for greater flexibility in working through trigonometric problems, like the conversion between \(\sin^{-1}\)\ and \(\cos^{-1}\)\ in this exercise.

Simplifying Expressions

Simplifying expressions is a key part of solving trigonometric equations. The primary aim is to reduce the complexity of an expression without changing its value. In this exercise, the simplification involves multiple steps.
First, compute individual components such as \(\sqrt{1-x^2}\) and \(\sqrt{1-y^2}\). These components are critical for substituting back into the inverse sine difference formula. The calculation simplifies to something manageable:

• Calculate the necessary square roots.
• Use multiplication to create a single fraction.
• Subtract within a common denominator to find \(\frac{33}{65}\)\.

The act of substitution and algebraic manipulation ultimately brings us to the simplified and correct form of the expression, which aids in identifying the correct answer choice.

Angles and Their Measures

Understanding angles and their measures is crucial in trigonometry, especially when distinguishing between functions like sine and cosine. An angle's measure can significantly impact these functions, often relating them in meaningful ways.
For inverse trigonometric functions, the angles are expressed within specific ranges. These functions allow us to determine angles from values of trigonometric functions, such as sine or cosine.

• In the case of arcsine \(\sin^{-1}\), the principal range is from \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\).
• Different functions have different interval and periodic behaviors.

Understanding these ranges and measures enables us to conclude that \( \pi - \cos^{-1}\left(\frac{33}{65}\right)\) is indeed equivalent to \( \sin^{-1}\left(\frac{33}{65}\right)\), establishing the right interpretation in the answer choices.