Problem 18 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

\(\tan \left(\frac{\pi}{4}+\frac{1}{2} \cos ^{-1} x\right)+\tan \left(\frac{\pi}{4}-\frac{1}{2} \cos ^{-1} x\right)\) \(x \neq 0\), is equal to (a) \(x\) (b) \(2 x\) (c) \(\frac{2}{x}\) (d) \(\frac{x}{2}\)

Step-by-Step Solution

Answer

The expression \(\tan \left(\frac{\pi}{4}+\frac{1}{2} \cos ^{-1} x\right) + \tan \left(\frac{\pi}{4}-\frac{1}{2} \cos ^{-1} x\right)\) simplifies to \(\frac{2}{x}\)

Step 1: Rewrite using the Tangent of Sum and Difference of angles formula

We have:\n\(\tan \left(\frac{\pi}{4}+\frac{1}{2} \cos ^{-1} x\right) + \tan \left(\frac{\pi}{4}-\frac{1}{2} \cos ^{-1} x\right) = \frac{\tan \frac{\pi}{4} + \tan \frac{1}{2}\cos^{-1}x}{1 - \tan \frac{\pi}{4} \tan \frac{1}{2}\cos^{-1}x} + \frac{\tan \frac{\pi}{4} - \tan \frac{1}{2}\cos^{-1}x}{1 + \tan \frac{\pi}{4} \tan \frac{1}{2}\cos^{-1}x}\)

Step 2: Simplify using Trigonometric Identities

As \(\tan \frac{\pi}{4} = 1\) and \(\cos^{-1}x = 2\cos^{-1}\sqrt{\frac{1 + \cos x}{2}}\), inserting these identities in the simplified expression we get:\n\(\frac{1 + \tan \frac{1}{2}\cos^{-1}x}{1 - \tan \frac{1}{2}\cos^{-1}x} + \frac{1 - \tan \frac{1}{2}\cos^{-1}x}{1 + \tan \frac{1}{2}\cos^{-1}x}\)\n further simplifying it yields to \(\frac{2}{\tan \frac{1}{2}\cos^{-1}x}\).

Step 3: Replace and observe

Now replacing \(cos^{-1}x\) with \(y\), then \(x = cos y\) and as \(y\) ranges from 0 to \(\pi\), x ranges from 1 to -1. Therefore as \(x\) changes from 1 to -1, \(\frac{1}{2} cos^{-1}x\) changes from 0 to \(\frac{\pi}{2}\) indicating that \(\tan \frac{1}{2}\cos^{-1}x = x\). So the expression \(\frac{2}{\tan \frac{1}{2}\cos^{-1}x}\) simplifies to \(\frac{2}{x}\)

Key Concepts

Tangent of Sum and Difference Formulas

When we encounter an expression involving the tangent of the sum or difference of two angles, we can decompose it into a more manageable form. The core formulas here are:\begin{align*}\tan(A + B) &= \frac{\tan A + \tan B}{1 - \tan A \tan B},\tan(A - B) &= \frac{\tan A - \tan B}{1 + \tan A \tan B}.\end{align*}Applying these to our exercise, where the angles are \(\frac{\pi}{4} + \frac{1}{2} \cos^{-1} x\) and \(\frac{\pi}{4} - \frac{1}{2} \cos^{-1} x\), simplifies the original complex expression into forms involving basic arithmetic on tangents of individual angles. This breakdown is crucial as it paves the way for further simplification using other trigonometric identities and eventually leading to a solution. With practice, students can master these formulas to solve complex trigonometric expressions, a skill particularly useful for IIT JEE aspirants.

Inverse Trigonometric Functions

Inverse trigonometric functions are a key part of high school and competitive exams like IIT JEE trigonometry. Notably, \(\cos^{-1}x\), which represents the angle whose cosine is x, plays a pivotal role in our solution. Inverse functions are useful for deducing angles when the trigonometric ratios are known. For instance:

• \(\sin^{-1}x\) defines the unique angle in \([ -\frac{\pi}{2}, \frac{\pi}{2} ]\) whose sine is x.
• \(\tan^{-1}x\) defines the angle in \([ -\frac{\pi}{2}, \frac{\pi}{2} ]\) whose tangent is x.

Combining these functions with other trigonometric identities, such as the half-angle formulas, allows for the simplification of seemingly complex expressions. The ability to manipulate these functions is essential for solving trigonometric equations, a fundamental aspect of advanced mathematics which is frequently tested in the IIT JEE.

Trigonometric Simplification

Trigonometric simplification can transform an intimidating trigonometric expression into something far less daunting. Essential to this process is the ability to recognize and adeptly apply basic trigonometric identities. Trigonometric simplification often involves strategies such as:

• Recognizing patterns that match known identities.
• Converting complex expressions into forms involving the sine and cosine ratios which are more intuitive to simplify.
• Utilizing inverse trigonometric properties and relationships between the functions.

In the exercise, the use of the identities like \(\tan \frac{\pi}{4} = 1\) is a strategic move that enabled the simplification. Moreover, understanding that \(\cos^{-1}x\) can be transformed using half-angle formulas propelled the simplification process further. This sophistication in approach not only leads to a refined solution but also lays a firm foundation for analytical thinking, an invaluable skill in mathematical problem-solving, especially in competitive exams like the IIT JEE. By familiarizing themselves with these techniques, students can approach their trigonometry problems with greater confidence and efficiency.