Problem 22 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

Solve for \(x\) : \(2 \tan ^{-1} x=\cos ^{-1}\left(\frac{1-a^{2}}{1+a^{2}}\right)-\cos ^{-1}\left(\frac{1-b^{2}}{1+b^{2}}\right)\), \(a>0, b>0 .\)

Step-by-Step Solution

Answer

After a careful step-by-step approach incorporating necessary transformations and simplifications, it has been determined that \(x = \frac{\sqrt{\frac{a^{2}-b^{2}}{1-a^{2}-b^{2}}} + \sqrt{\frac{b^{2}-a^{2}}{1-b^{2}+a^{2}}}} {1 - \sqrt{\frac{a^{2}-b^{2}}{1-a^{2}-b^{2}}} \sqrt{\frac{b^{2}-a^{2}}{1-b^{2}+a^{2}}}}\)

Step 1: Apply Inverse Cosine Transformation

Firstly, identify the inverse cosine function expressions and transform them into the tangent form using the identity \(\cos ^{-1} x = \frac{\pi}{2} - \tan ^{-1} \sqrt{\frac{1-x}{1+x}}\), we get: \(2 \tan ^{-1} x = \frac{\pi}{2} - \tan ^{-1} \sqrt{\frac{1-\frac{1-a^{2}}{1+a^{2}}} {1+\frac{1-a^{2}}{1+a^{2}}}} - \frac{\pi}{2} + \tan ^{-1} \sqrt{\frac{1-\frac{1-b^{2}}{1+b^{2}}}{1+\frac{1-b^{2}}{1+b^{2}}}}\)

Step 2: Simplify the equation

On simplifying the above equation, the terms \(\frac{\pi}{2}\) get cancelled, hence the equation becomes: \(2 \tan^{-1}x = \tan^{-1} \sqrt{\frac{a^{2}-b^{2}}{1-a^{2}-b^{2}}} - \tan^{-1} \sqrt{\frac{b^{2}-a^{2}}{1-b^{2}+a^{2}}}\)

Step 3: Apply Addition Formula for Inverse Tangent

Applying the addition formula, which states that \(\tan^{-1}x + \tan^{-1}y = \tan^{-1}\frac{x+y}{1-xy}\) if \(xy<1\), results in: \(2 \tan^{-1}x = \tan^{-1}\left(\frac{\sqrt{\frac{a^{2}-b^{2}}{1-a^{2}-b^{2}}} + \sqrt{\frac{b^{2}-a^{2}}{1-b^{2}+a^{2}}}} {1- \sqrt{\frac{a^{2}-b^{2}}{1-a^{2}-b^{2}}} \sqrt{\frac{b^{2}-a^{2}}{1-b^{2}+a^{2}}}}\right)\)

Step 4: Isolate x

Now remove the \(\tan^{-1}\) on both sides. By dividing the two sides by 2 we isolate x to get: \(x = \frac{\sqrt{\frac{a^{2}-b^{2}}{1-a^{2}-b^{2}}} + \sqrt{\frac{b^{2}-a^{2}}{1-b^{2}+a^{2}}}} {1 - \sqrt{\frac{a^{2}-b^{2}}{1-a^{2}-b^{2}}} \sqrt{\frac{b^{2}-a^{2}}{1-b^{2}+a^{2}}}}\)

Key Concepts

Tangent Form

The tangent form is used to express trigonometric functions in terms of the tangent of half an angle. This approach is particularly useful for simplifying complex trigonometric expressions. By converting inverse trigonometric functions like inverse cosine into tangent form, we can make calculations more straightforward.

Consider the formula: \[\cos^{-1}x = \frac{\pi}{2} - \tan^{-1} \sqrt{\frac{1-x}{1+x}}\] This identity allows us to transform the inverse cosine term into an equivalent arctangent expression. It helps to solve equations where multiple trigonometric identities are involved. The conversion simplifies the equation by reducing it to one type of function, making it easier to apply additional formulas such as the addition or subtraction formulas.

Addition Formula

The addition formula for inverse tangent is a crucial part of simplifying equations involving multiple inverse tangent terms. The formula is expressed as: \[\tan^{-1}x + \tan^{-1}y = \tan^{-1}\frac{x+y}{1-xy}\] provided that \(xy < 1\).

This formula relates the sum of two inverse tangent angles to a single inverse tangent of a combined argument. It is particularly helpful in solving problems where such expressions arise, as it reduces the complexity by consolidating multiple terms into one.

• Make sure to check the condition \(xy<1\) before applying this formula, as it ensures that the expression is valid.
• Using the formula simplifies calculations and often leads to a direct path to isolating the variable in question.

This was applied in our solution to manage two separate inverse tangent terms and helped isolate \(x\) effectively by bringing them into a simpler form.

Inverse Cosine Transformation

Inverse cosine transformation is a method that helps convert \( \cos^{-1} \) terms into other forms, like \( \tan^{-1} \). This transformation, as seen in our solution, leverages identities to rewrite equations in more manageable forms. For example, converting \( \cos^{-1}\left(\frac{1-a^{2}}{1+a^{2}}\right)\) and \(\cos^{-1}\left(\frac{1-b^{2}}{1+b^{2}}\right)\) into tangent forms gives us a pathway to apply addition formulas efficiently.

By transforming these inverse cosine expressions, we simplify their manipulation and interaction with other terms.

• This is particularly useful in equations involving several inverse trigonometric functions and simplifies the path to a solution.
• Switching between inverse cosine and inverse tangent is often necessary to unlock analytical approaches that aren't feasible with direct inverse cosine methods alone.

Employing this technique ensures flexibility in mathematical processes, leading to successful solutions of intricate trigonometric identities.