Problem 85 - A Complete Resource Book in Mathematics for JEE Main

Question

The value of \(\cot \left(\operatorname{cosec}^{-1} \frac{5}{3}+\tan ^{-1} \frac{2}{3}\right)\) is (A) \(\frac{6}{17}\) (B) \(\frac{3}{17}\) (C) \(\frac{4}{17}\) (D) \(\frac{5}{17}\)

Step-by-Step Solution

Answer

The value of \( \cot \left(\operatorname{cosec}^{-1} \frac{5}{3} + \tan^{-1} \frac{2}{3}\right) \) is \( \frac{6}{17} \).

Step 1: Define the expressions

Let \( \theta = \csc^{-1} \left( \frac{5}{3} \right) \) and \( \phi = \tan^{-1} \left( \frac{2}{3} \right) \). The goal is to find \( \cot(\theta + \phi) \).

Step 2: Express \(\theta\) in terms of sine

For \( \theta = \csc^{-1} \left( \frac{5}{3} \right) \), we have \( \csc\theta = \frac{5}{3} \). This implies that \( \sin\theta = \frac{3}{5} \).

Step 3: Find \(\cot\theta\) and \(\tan\theta\)

Calculate \( \cot\theta = \frac{\cos\theta}{\sin\theta} \). Using \( \sin\theta = \frac{3}{5} \), we have \( \cos\theta = \sqrt{1 - \sin^2\theta} = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \frac{4}{5} \). Thus, \( \cot\theta = \frac{4}{3} \).

Step 4: Express \(\phi\) in terms of tangent

For \( \phi = \tan^{-1} \left( \frac{2}{3} \right) \), \( \tan\phi = \frac{2}{3} \).

Step 5: Use the cotangent addition formula

The formula \( \cot(\theta + \phi) = \frac{\cot\theta \cdot \cot\phi - 1}{\cot\theta + \cot\phi} \) is used to find \( \cot(\theta + \phi) \).

Step 6: Calculate \(\cot\phi\)

Since \( \tan\phi = \frac{2}{3} \), we have \( \cot\phi = \frac{1}{\tan\phi} = \frac{3}{2} \).

Step 7: Substitute into the cotangent addition formula

Substitute \( \cot\theta = \frac{4}{3} \) and \( \cot\phi = \frac{3}{2} \) into the formula: \[ \cot(\theta + \phi) = \frac{\left(\frac{4}{3}\right) \left(\frac{3}{2}\right) - 1}{\frac{4}{3} + \frac{3}{2}} = \frac{2 - 1}{\frac{8 + 9}{6}} = \frac{1}{\frac{17}{6}} = \frac{6}{17}. \]

Step 8: Determine the answer

The value of \( \cot(\theta + \phi) \) is \( \frac{6}{17} \), corresponding to option (A).

Key Concepts

Inverse Trigonometric Functions

Understanding inverse trigonometric functions can help in unraveling complex trigonometric equations. These functions serve the purpose of finding an angle when given a trigonometric ratio. For example,

• The inverse cosecant function, denoted as \( \csc^{-1} \), helps us find the angle whose cosecant value is known.
• Similarly, the inverse tangent function, \( \tan^{-1} \), finds the angle for a given tangent.

If \( \theta = \csc^{-1} \left( \frac{5}{3} \right) \), it indicates that \( \csc\theta = \frac{5}{3} \). From this, you can deduce that \( \sin\theta = \frac{3}{5} \). Recognizing that \( \phi = \tan^{-1} \left( \frac{2}{3} \right) \) means \( \tan\phi = \frac{2}{3} \) is similarly useful. These functions are crucial in simplifying trigonometric expressions and finding relevant angles that work within those expressions.

Cotangent Addition Formula

This formula is a handy tool for combining angles in trigonometry. To add the cotangent of two angles, \( \theta \) and \( \phi \), the cotangent addition formula is used:\[\cot(\theta + \phi) = \frac{\cot\theta \cdot \cot\phi - 1}{\cot\theta + \cot\phi}.\]This result comes from manipulating the tangent addition formula. It stems from identities in trigonometry that relate sine, cosine, and tangent with cotangent.In the exercise, we needed this formula to compute the value of \( \cot(\theta + \phi) \) efficiently.

• By substituting in \( \cot\theta = \frac{4}{3} \) and \( \cot\phi = \frac{3}{2} \), the expression simplifies to \( \cot(\theta + \phi) = \frac{6}{17} \).

Understanding how this formula is derived and applied can significantly simplify solving related trigonometric equations in both theoretical and practical contexts.

Trigonometric Identities

These identities are like building blocks for solving trigonometric problems. They allow the expression of trigonometric functions in terms of one another. For example,

• Pythagorean identities, such as \( \sin^2 x + \cos^2 x = 1 \), are profound in establishing relationships between the functions.
• Reciprocal identities like \( \csc x = \frac{1}{\sin x} \) and \( \cot x = \frac{1}{\tan x} \) are often used to simplify expressions.

In the original exercise, we transformed \( \csc^{-1} \left( \frac{5}{3} \right) \) into \( \sin\theta = \frac{3}{5} \) using a reciprocal identity. Then, we calculated \( \cos\theta = \frac{4}{5} \) based on the Pythagorean identity. This led us to \( \cot\theta = \frac{4}{3} \).Mastering these identities equips students to break down complex problems into manageable components, offering multiple pathways to the solution.