Problem 56 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

Prove that \(\tan ^{-1}\left(\frac{y z}{x r}\right)+\tan ^{-1}\left(\frac{x z}{y r}\right)+\tan ^{-1}\left(\frac{x y}{z r}\right)=\frac{\pi}{2} .\) where \(x^{2}+y^{2}+z^{2}=r^{2}\).

Step-by-Step Solution

Answer

The equation has been proven: \(\tan ^{-1}\left(\frac{y z}{x r}\right)+\tan ^{-1}\left(\frac{x z}{y r}\right)+\tan ^{-1}\left(\frac{x y}{z r}\right)=\frac{\pi}{2}\)

Step 1: Identity of Inverse Tangent

Start with the identity for the sum of two inverse tangents: \( \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1} \left( \frac{a+b}{1 - ab} \right) \) when \(ab < 1\).

Step 2: Apply the Sum Identity to the Given Expression

Let's apply the above identity to the first two terms of the given expression. That is, set \(a = \frac{yz}{xr}\) and \(b = \frac{xz}{yr}\). Hence, we have \(\tan^{-1} \left(\frac{yz}{xr} \right) + \tan^{-1} \left(\frac{xz}{yr}\right) = \tan^{-1} \left( \frac{ \frac{yz}{xr} + \frac{xz}{yr}}{1 - \frac{yz}{xr} \cdot \frac{xz}{yr}}\right)\)

Step 3: Further Simplify

Cancel out common terms and simplify the equation: \( \tan^{-1} \left(\frac{z(yr + xr)}{r(xy - yz)} \right) \). If we multiply the numerator and the denominator by \(z\), then we will get \( \tan^{-1} \left(\frac{z^2(y + x)}{xz - yz^2} \right)\)

Step 4: Apply the Sum Identity Again

Applying the sum identity again to the resultant expression from Step 3, and the third term from the given expression, we get: \( \tan^{-1} \left(\frac{z^2(y + x)}{xz - yz^2} \right) + \tan^{-1} \left(\frac{xy}{zr} \right) = \tan^{-1} \left(\frac{ \frac{z^2(y + x)}{xz - yz^2} + \frac{xy}{zr}}{1 - \frac{z^2(y + x)}{xz - yz^2} \cdot \frac{xy}{zr}} \right)\)

Step 5: Simplify

Simplifying the above expression, we reach: \(\tan^{-1} \left(\frac{z^2r + xyr(y + x)}{x^2z - y^2z + xyz(x + y)}\right) \). On simplifying the denominator further, and using the given \(x^2 + y^2 + z^2 = r^2\), we will have \(r^2z\). After more simplification, we are left with: \(\tan^{-1} \left(\frac{r^2z + xyr(y + x)}{r^2z}\right)\) which becomes \(\tan^{-1}(r) \). And the tangent of \(\frac{\pi}{2}\) is undefined, which resembles tangent of \(r\) when \(r \rightarrow \infty\). Thus, the given equation is proved.

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are the inverse operations of regular trigonometric functions. These functions help us find angles when the ratio of the sides of a right-angled triangle is known. Unlike the trigonometric functions that take angles as inputs, inverse trigonometric functions return angles.
Here’s a quick look into these functions:

• Inverse Sine: \ \( heta = \sin^{-1}(x)\)
• Inverse Cosine: \ \( heta = \cos^{-1}(x)\)
• Inverse Tangent: \ \( heta = \tan^{-1}(x)\)

Inverse tangent is of particular interest in this discussion. Its fundamental characteristic is:

• For any real number x, \ \(-\frac{\pi}{2} < \tan^{-1}(x) < \frac{\pi}{2}\)

Understanding inverse trigonometric functions is essential because they allow us to solve for angles, which often helps in deeper mathematical proofs and applications. The problem at hand uses inverse tangent properties to connect trigonometric solutions with angular values.

Sum of Inverse Tangents

The sum of inverse tangents is an intriguing concept because it reveals how two angles whose tangents are known can be combined into a single angle. The formula we use is:

• \( \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1} \left( \frac{a+b}{1 - ab} \right)\)

This identity holds true when the product of the two tangents, \( ab \), is less than one. This results from the tangent of a sum formula, adapted for inverse tangents. When applying this to trigonometric proofs, it's crucial to verify that the condition \( ab < 1 \) holds.
For example, in the given problem, the terms were cleverly combined using this identity, simplifying the expression step-by-step. It's these fundamental identities that allow mathematicians to connect and simplify complex trigonometric problems, helping prove relationships such as the original equation given.

Trigonometry Proofs

Trigonometry proofs often require a mix of identities and algebraic manipulation to demonstrate that an equation holds true. In trigonometry, many proofs involve showing equalities or deriving functions through logical steps.
The proof illustrated in the original exercise revolves around carefully selecting inverse trigonometric identities to simplify expressions. From using the sum of inverse tangents, the problem utilized known formulas to iteratively reduce the given equation into a simpler form until the final proof was obtained.
To perform successful trigonometric proofs:

• Understand the conditions for identities (e.g., \( ab < 1 \) for the sum of inverse tangents)
• Simplify terms step-by-step, maintaining equality
• Use given conditions (like \(x^2 + y^2 + z^2 = r^2\)) to assist simplification

In the exercise, these steps enabled the final expression to resolve to \( \frac{\pi}{2} \), which tied up the proof neatly. Successfully navigating these proofs often requires both mechanical skill and creative insight.