Question

If \(\tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{5}\right)+\tan ^{-1}\left(\frac{1}{7}\right)=\tan ^{-1} \mathrm{x}\), then the value of \(\mathrm{x}\) is (a) \(\frac{3}{4}\) (b) \(\frac{7}{9}\) (c) \(\frac{12}{13}\) (d) \(\frac{13}{12}\)

Short answer

Based on the given solution, the value of x for which the sum of the three inverse tangent functions equals the inverse tangent of x is (c) x = 12/13.

Step-by-step solution

Split the sum into two parts

We will split the given sum of inverse tangent functions into two parts: \(\tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{5}\right)+\tan ^{-1}\left(\frac{1}{7}\right) = \left[\tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{5}\right)\right] + \tan ^{-1}\left(\frac{1}{7}\right)\)

Apply the sum-to-product formula for the first two terms

Now we will apply the sum-to-product formula for the sum of the first two inverse tangent functions: \(\tan\left[\tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{5}\right)\right] = \frac{\tan(\tan ^{-1}\left(\frac{1}{3}\right)) + \tan(\tan ^{-1}\left(\frac{1}{5}\right))}{1 - \tan(\tan ^{-1}\left(\frac{1}{3}\right)) \tan(\tan ^{-1}\left(\frac{1}{5}\right))}\) \(= \frac{\frac{1}{3} + \frac{1}{5}}{1 - \frac{1}{3}.\frac{1}{5}}\) Simplify the expression: \(= \frac{\frac{2}{5}}{\frac{12}{15}} = \frac{13}{12}\) So, we have: \(\tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{5}\right) = \tan^{-1}\left(\frac{13}{12}\right)\) Now our expression is: \(\tan^{-1}\left(\frac{13}{12}\right) + \tan ^{-1}\left(\frac{1}{7}\right) = \tan^{-1}(x)\)

Apply the sum-to-product formula again

Now, we will apply the sum-to-product formula again for the sum of the remaining two inverse tangent functions: \(\tan\left[\tan^{-1}\left(\frac{13}{12}\right) + \tan ^{-1}\left(\frac{1}{7}\right)\right] = \frac{\tan(\tan^{-1}\left(\frac{13}{12}\right)) + \tan(\tan ^{-1}\left(\frac{1}{7}\right))}{1 - \tan(\tan^{-1}\left(\frac{13}{12}\right)) \tan(\tan ^{-1}\left(\frac{1}{7}\right))}\) \(= \frac{\frac{13}{12} + \frac{1}{7}}{1 - \frac{13}{12}.\frac{1}{7}}\) Simplify the expression: \(= \frac{\frac{27}{28}}{\frac{11}{12}} = \frac{12}{13}\) Therefore, we have: \(\tan^{-1}\left(\frac{13}{12}\right) + \tan ^{-1}\left(\frac{1}{7}\right) = \tan^{-1}(x)\)

Find the value of x

Comparing the obtained expression with the given expression, we can deduce that: \(\tan^{-1}(x) = \tan^{-1}\left(\frac{12}{13}\right)\) So, the value of x is \(\frac{12}{13}\). Hence, the correct answer is (c) \(\frac{12}{13}\).

Key concepts

Sum of Inverse Tangents

Understanding the sum of inverse tangents is crucial not only in trigonometry but also in various applications such as calculus and geometry. The sum of inverse tangent functions, such as \(\tan ^{-1}(a) + \tan ^{-1}(b)\), often appears in mathematical problems and can be simplified using trigonometric identities.

For instance, the expression \(\tan ^{-1}(\frac{1}{3}) + \tan ^{-1}(\frac{1}{5})\) can be interpreted as the sum of the angles whose tangent values are \(\frac{1}{3}\) and \(\frac{1}{5}\), respectively. To solve for the sum, one typically applies a sum-to-product formula or other trigonometric identities to simplify the expression. The importance of recognizing these types of terms and knowing how to work with them is essential for problem-solving in trigonometry-focused exams like IIT-JEE.

Sum-to-Product Formula

The sum-to-product formula is a powerful tool in trigonometry used to convert the sum or difference of trigonometric functions into a product of trigonometric functions. It helps in simplifying the computation and solving trigonometric equations efficiently.

When applied to inverse tangent sums, for instance, the sum \(\tan ^{-1}(a) + \tan ^{-1}(b)\) can be simplified using the identity: \(\tan(\tan ^{-1}(a) + \tan ^{-1}(b)) = \frac{a+b}{1-ab}\), provided the denominator does not equal zero. This transformation is partly why the sum-to-product formula is particularly useful in both analytical and computational contexts, simplifying complex expressions into more manageable ones.

IIT-JEE Mathematics Problems

IIT-JEE is an advanced engineering entrance examination in India, known for its challenging mathematics problems. These problems often include complex trigonometric equations, inverse functions, and their properties.

Mathematics problems in IIT-JEE not only test a student's understanding of concepts but also their application and problem-solving skills. Trigonometry, especially topics like inverse trigonometric functions and their sums, plays a significant role due to its extensive use in calculus and algebra. Grasping these concepts is crucial for success in such competitive exams, where every mark can make a difference in the outcome.

Trigonometric Equation Solving

Solving trigonometric equations is about finding the values of the variables that satisfy the given trigonometric expression. These equations can range from basic to highly complex, requiring a strong grasp of trigonometric identities and formulas.

To tackle these equations, one must be familiar with the fundamental trigonometric functions and their inverses, transformation formulas like the sum-to-product, and methods for simplifying and solving expressions. Mastery of these techniques is not only essential for academic purposes but also for real-world applications where trigonometric equations are used to model phenomena.