Question

When \(\mathrm{y}>\mathrm{x}>0, \cot ^{-1} \mathrm{x}-\cot ^{-1} \mathrm{y}\) is equal to (a) \(\cot ^{-1}\left(\frac{1-x y}{y+x}\right)\) (b) \(\cot ^{-1}\left(\frac{y+x}{1+x y}\right)\) (c) \(\cot ^{-1}\left(\frac{1+x y}{y-x}\right)\) (d) \(\cot ^{-1}\left(\frac{1-x y}{y-x}\right)\)

Short answer

Question: Prove that if \(y>x>0\), then the expression \(\cot^{-1}\mathrm{x} - \cot^{-1}\mathrm{y}\) is equal to: Options: (a) \(\cot^{-1}\left(\frac{1-x \cdot y}{y+x}\right)\) (b) \(\cot^{-1}\left(\frac{y+x}{1+x \cdot y}\right)\) (c) \(\cot^{-1}\left(\frac{1+x \cdot y}{y-x}\right)\) (d) \(\cot^{-1}\left(\frac{1-x \cdot y}{y-x}\right)\) Answer: (c) \(\cot^{-1}\left(\frac{1+x \cdot y}{y-x}\right)\)

Step-by-step solution

Write down the given expression

We start with the expression: \(\cot^{-1}\mathrm{x} - \cot^{-1}\mathrm{y}\).

Express in terms of tan and apply the formula

We can express the expression in terms of tangent function using the property \(\cot^{-1}x = \tan^{-1}(\frac{1}{x})\). Therefore, our expression becomes: \(\tan^{-1}(\frac{1}{x}) - \tan^{-1}(\frac{1}{y})\). Now, we will apply the formula for the difference of two inverse tangent functions: \(\tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1}(\frac{a-b}{1+ab})\) Here, \(a = \frac{1}{x}\) and \(b = \frac{1}{y}\). Plugging the values, we get: \(\tan^{-1}(\frac{1}{x}) - \tan^{-1}(\frac{1}{y}) = \tan^{-1}\left(\frac{\frac{1}{x}-\frac{1}{y}}{1+\frac{1}{x}\cdot\frac{1}{y}}\right)\).

Simplify the expression

Simplify the expression by finding a common denominator and further rearranging the terms: \(\tan^{-1}\left(\frac{\frac{y-x}{x \cdot y}}{1+\frac{1}{x \cdot y}}\right) = \tan^{-1}\left(\frac{y - x}{x \cdot y + 1}\right)\)

Express in terms of cot

Now we will switch back to cotangent function using the property \(\tan^{-1}x = \cot^{-1}(\frac{1}{x})\). Therefore, our simplified expression becomes: \(\cot^{-1}\left(\frac{x \cdot y + 1}{y - x}\right)\)

Compare with the given options

Compare the simplified expression with the four options provided: (a) \(\cot^{-1}\left(\frac{1-x \cdot y}{y+x}\right)\), not equal (b) \(\cot^{-1}\left(\frac{y+x}{1+x \cdot y}\right)\), not equal (c) \(\cot^{-1}\left(\frac{1+x \cdot y}{y-x}\right)\), this is the correct option (d) \(\cot^{-1}\left(\frac{1-x \cdot y}{y-x}\right)\), not equal Hence, the correct form is (c) \(\cot^{-1}\left(\frac{1+x \cdot y}{y-x}\right)\).

Key concepts

Cot Inverse Formula

Understanding the cot inverse formula is crucial for students, as it provides a way to represent angles when given a specific cotangent value. The function \( \cot^{-1}(x) \) denotes the angle whose cotangent is \( x \). In trigonometry, cotangent is the reciprocal of the tangent function, and the inverse cotangent function can be linked to the inverse tangent function as follows: \( \cot^{-1}(x) = \tan^{-1}(\frac{1}{x}) \).

This relationship is particularly handy when simplifying expressions involving cotangent inverses, as demonstrated in the exercise. By substituting the cotangent inverse with the tangent inverse, we can apply the tan inverse subtraction formula to find the difference between two angles, which introduces the next concept in our discussion.

Tan Inverse Subtraction Formula

The tan inverse subtraction formula plays a pivotal role in simplifying problems involving the difference between two angles. According to this principle, \( \tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1}(\frac{a-b}{1+ab}) \).

This property allows us to merge two inverse tangent functions into a single expression, which is simpler to address. To apply this formula effectively, expressions must first be converted into a suitable form, often necessitating rendering expressions with cotangents into their tangent counterparts. In the provided step-by-step solution, this subtraction formula transformed the complex inverse cotangent difference into a simplified version expressed as a single inverse cotangent.

Inverse Trigonometric Functions Simplification

Simplification of inverse trigonometric functions is an essential skill, especially in complex calculations. The key to such simplifications lies in understanding and adeptly applying trigonometric identities and properties. In the context of the given solution, the expression cleanup involved finding common denominators and restructuring terms.

These techniques often include arithmetical manipulation, such as expressing compound fractions in their simplest form, which then can be easily rewritten using inverse trigonometric relationships like \( \tan^{-1}(x) = \cot^{-1}(\frac{1}{x}) \). The goal is to ultimately match the simplified expression with possible solutions or further simplify them to yield a final, more straightforward expression.

IIT JEE Mathematics Problems

IIT JEE, the Indian Institutes of Technology Joint Entrance Examination, features a variety of challenging mathematics problems, including those on inverse trigonometric functions. Mastery over concepts like the cot inverse formula and tan inverse subtraction formula is vital to tackle these problems successfully.

Many IIT JEE aspirants often find these topics tricky due to their abstract nature. However, approaching the problems methodically by breaking them down into steps, as shown in the exercise solution, can make them manageable. Utilizing problem-solving strategies such as rewriting expressions and simplifying can lead to successful results on the exam. Students preparing for IIT JEE must practice rigorously with a wide range of problems to develop a strong foundation in these and other complex mathematical concepts.