Problem 26 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

The values of \(x\) satisfying \(\tan \left(\sec ^{-1} x\right)=\sin \left(\cos ^{-1}\left(\frac{1}{\sqrt{5}}\right)\right)\) is (a) \(\pm \frac{\sqrt{5}}{3}\) (b) \(\pm \frac{3}{\sqrt{5}}\) (c) \(\pm \frac{\sqrt{3}}{5}\) (d) \(\pm \frac{3}{5}\)

Step-by-Step Solution

Answer

The values of \(x\) that satisfy the given equation are \(\pm \frac{3\sqrt{5}}{5}\) (option b).

Step 1: Simplify the RHS

We first simplify \(\sin \left(\cos^{-1}\left(\frac{1}{\sqrt{5}}\right)\right)\). Based on the Pythagorean identity, we have \(\sin(y) = \sqrt{1 - \cos^2(y)}\). So, substituting \(\frac{1}{\sqrt{5}}\) for \(\cos(y)\), we have \(\sin(y) = \sqrt{1 - (\frac{1}{\sqrt{5}})^2} = \sqrt{1-\frac{1}{5}} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}} \). Meanwhile, we standardize this expression as \( \frac{2\sqrt{5}}{5} \) which implies \(\sin (\cos^{-1}(\frac{1}{\sqrt{5}})) = \frac{2\sqrt{5}}{5}\).

Step 2: Simplify the LHS

Next, we solve \(\tan \left(\sec ^{-1} x\right)\). Because \(\sec(y) = \frac{1}{\cos(y)}\), we can say \(\cos(y) = \frac{1}{x}\). Moreover, as the tangent function is \(\tan(y) = \frac{\sin(y)}{\cos(y)}\), we substitute \(\frac{1}{x}\) for \(\cos(y)\) and get \(\tan(y) = \frac{\sin(y)}{1/x} = x \sin(y)\). But, by the Pythagorean identity, \(\sin(y) = \sqrt{1 - \cos^2(y)} = \sqrt{1 - (\frac{1}{x})^2} = \frac{\sqrt{x^2 - 1}}{x}\). Therefore, \(\tan(\sec^{-1}(x)) = x \cdot \frac{\sqrt{x^2 - 1}}{x} = \sqrt{x^2 - 1}\).

Step 3: Solve for x

Setting the RHS equal to the LHS: \(\sqrt{x^2 - 1} = \frac{2\sqrt{5}}{5}\). To proceed, we can square both sides of the equation in order to eliminate the square root and facilitate the process of isolating \(x\), so we obtain \(x^2 - 1 = \frac{20}{25} = \frac{4}{5}\). Adding 1 to both sides, we find \(x^2 = 1 + \frac{4}{5} = \frac{9}{5}\). Finally, taking the square root gives the values of \(x\) as \(\pm \frac{3}{\sqrt{5}} = \pm \sqrt{\frac{9}{5}}\). Simplifying these results \(x = \pm \frac{3\sqrt{5}}{5}\), which is choice (b) in the given options.

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are essential when we want to find an angle with a specific trigonometric value. These functions help us work backward from a ratio to an angle, which can then be used in further calculations.

In this exercise, we encounter expressions such as \( \sec^{-1}(x) \) and \( \cos^{-1}( \frac{1}{\sqrt{5}} ) \). Let's break them down:

• The inverse secant function, \( \sec^{-1}(x) \), returns an angle \( y \) such that \( \sec(y) = x \).
• For the inverse cosine, \( \cos^{-1}( \frac{1}{\sqrt{5}} ) \), it returns the angle \( y \) where \( \cos(y) = \frac{1}{\sqrt{5}} \).

Inverse functions are crucial for translating between trigonometric ratios and angles in solving equations like this. Understanding these concepts allows us to tackle more complex trigonometric relationships.

Pythagorean Identity

The Pythagorean identity is a fundamental relation in trigonometry that connects the sine and cosine functions: \( \sin^2(y) + \cos^2(y) = 1 \). This identity allows us to express one trigonometric function in terms of another.

In this problem, the identity is used to express \( \sin(y) \) in terms of \( \cos(y) \), specifically \( \sin(\cos^{-1}( \frac{1}{\sqrt{5}} )) \). This is done by calculating:

• \( \sin(y) = \sqrt{1 - \cos^2(y)} \)
• Substituting \( \cos(y) = \frac{1}{\sqrt{5}} \), we have \( \sin(y) = \sqrt{1 - (\frac{1}{\sqrt{5}})^2} = \frac{2}{\sqrt{5}} \)

The identity simplifies complex expressions and is vital for solving trigonometric equations. Mastery of this tool sets a strong foundation for understanding trigonometric relationships.

Simplifying Trigonometric Expressions

Simplifying trigonometric expressions involves reducing complex trigonometric terms to simpler forms. This process is essential for solving equations and finding identities in trigonometry.

In the solution, both sides of the given equation are simplified:

• For the right-hand side, we use the Pythagorean identity to simplify \( \sin(\cos^{-1}( \frac{1}{\sqrt{5}} )) \) to \( \frac{2\sqrt{5}}{5} \).
• For the left-hand side, the expression \( \tan(\sec^{-1}(x)) \) is simplified using \( \tan(y) = \frac{\sin(y)}{\cos(y)} \), resulting in \( \sqrt{x^2 - 1} \).

This simplification makes the equation easier to solve, especially when setting the simplified expressions equal to one another. By learning to simplify effectively, you'll gain the ability to handle varied and intricate trigonometric problems with ease.