Problem 81 - A Complete Resource Book in Mathematics for JEE Main

Question

\(\tan ^{-1}\left(\frac{1}{4}\right)+\tan ^{-1}\left(\frac{2}{9}\right)\) is equal to: (A) \(\frac{1}{2} \cos ^{-1}\left(\frac{3}{5}\right)\) (B) \(\frac{1}{2} \sin ^{-1}\left(\frac{3}{5}\right)\) (C) \(\frac{1}{2} \tan ^{-1}\left(\frac{3}{5}\right)\) (D) \(\tan ^{-1}\left(\frac{1}{2}\right)\)

Step-by-Step Solution

Answer

The answer is (D) \( \tan^{-1}\left(\frac{1}{2}\right) \).

Step 1: Review the Formula for Adding Inverse Tangents

The sum of two inverse tangent functions can be expressed using the formula:\[tan^{-1}(x) + tan^{-1}(y) = tan^{-1}\left(\frac{x + y}{1 - xy}\right),\]provided \(xy < 1\). In this case, check if this condition is satisfied for \( \frac{1}{4} \) and \( \frac{2}{9} \).

Step 2: Check the Condition xy < 1

Calculate the product of \( \frac{1}{4} \) and \( \frac{2}{9} \) to verify the condition.\[ \frac{1}{4} \times \frac{2}{9} = \frac{2}{36} = \frac{1}{18}, \]which is indeed less than 1, so the condition is satisfied.

Step 3: Apply the Formula

Now, substitute \( x = \frac{1}{4} \) and \( y = \frac{2}{9} \) into the formula:\[ tan^{-1}\left(\frac{1}{4} + \frac{2}{9}\right) + tan^{-1}\left(\frac{0}{1 - \frac{1}{4} \times \frac{2}{9}}\right) = tan^{-1}\left(\frac{\frac{17}{36}}{\frac{17}{18}}\right) = tan^{-1}\left(\frac{9}{17}\right). \]However, this calculation is wrong. Correctly we should have:\[ tan^{-1}\left(\frac{\frac{1}{4} + \frac{2}{9}}{1 - \frac{1}{4} \times \frac{2}{9}}\right) = tan^{-1}\left(\frac{9 + \frac{8}{9}}{\frac{17}{18}}\right) = tan^{-1}\left(\frac{17}{34}\right) = tan^{-1}\left(\frac{1}{2}\right). \]

Step 4: Verify the Result

Recognize that \( \tan^{-1}\left(\frac{1}{2}\right) \) can match option \( (D) \). Reassess any intermediate step if another check was needed, but here the process verifies as correct.

Key Concepts

Addition of Inverse Tangents

The addition of inverse tangents is a fascinating aspect of trigonometry that allows us to simplify complex expressions involving inverse functions. When you have two numbers, say, \( x \) and \( y \) and you want to find the sum of their inverse tangents, you can use the formula:\[ \tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x + y}{1 - xy}\right) \]This formula is applicable provided the product \( xy < 1 \), ensuring the result remains valid and real.
In our example, we apply this with \( x = \frac{1}{4} \) and \( y = \frac{2}{9} \), yielding:- First, check the condition: \( \frac{1}{4} \times \frac{2}{9} = \frac{1}{18} < 1 \), satisfying the requirement. - Then, substitute these values into the formula: \[ \tan^{-1}\left(\frac{\frac{1}{4} + \frac{2}{9}}{1 - \frac{1}{4} \times \frac{2}{9}}\right) \]This calculation simplifies to \( \tan^{-1}\left(\frac{1}{2}\right) \), showing how the original expression is streamlined using this approach.

Trigonometric Identities

Trigonometric identities form the backbone of many mathematical manipulations. These identities, such as those involving tangent, sine, and cosine functions, help simplify complex problems.
In our context, we primarily focus on the addition of inverse tangent identities. This identity helps convert a sum of inverse tangents into a single expression, making it easier to solve problems involving angles.

• **Tangent Addition Rules:** Utilize the identity for the sum of the tangent function, which is crucial in deriving the inverse tangent sum formula.
• **Simplification:** The identity \( \tan^{-1}\left(\frac{1}{2}\right) \) simplifies understanding as it directly corresponds with option (D) from the exercise, validating our approach.

Knowing these identities means you're better equipped to handle tricky trigonometric problems, particularly those in exams or mathematical competitions.

JEE Mathematics

Understanding inverse trigonometric functions is vital not only in pure mathematics but also in competitive exams like the JEE (Joint Entrance Examination).
The JEE examination is a significant hurdle for many aspiring engineering students in India. Mastering core concepts, such as trigonometric identities and inverse functions, gives students a much-needed edge.

• **Effective Preparation:** The thorough understanding of topics like addition of inverse tangents ensures a smoother solving process during the JEE.
• **Problem-Solving Skills:** Many problems in JEE revolve around mathematical manipulation and problem-solving skills, as seen in our example.

By practicing these concepts, JEE aspirants can enhance their aptitude, ensuring they are well-prepared for various questions concerning trigonometry.