Question

In \(\Delta ABC\) if \(\frac{\tan(A-B)}{\tan A} + \frac{\sin^2 C}{\sin^2 A} = 1\) where \(A, B, C \in (0, \frac{\pi}{2})\) then

• A. \(\tan A, \tan B, \tan C\) are in A.P.
• B. \(\tan A, \tan C, \tan B\) are in A.P.
• C. \(\tan A, \tan B, \tan C\) are in G.P.
• D. \(\tan A, \tan C, \tan B\) are in G.P.

Hint:

In triangle problems, replace \( C \) with \( 180^\circ - (A+B) \) to reduce the problem to two variables.

Answer 1

The Correct Option is D

To solve the given problem, we need to analyze the existing equation and the required condition. The equation provided is:

\[\frac{\tan(A-B)}{\tan A} + \frac{\sin^2 C}{\sin^2 A} = 1\]

We want to verify or deduce the relationship among \(\tan A, \tan B, \tan C\). The provided solution which corresponds with the statement \(\tan A, \tan C, \tan B\) being in a Geometric Progression (G.P.) is indeed true. Let's break it down step-by-step:

1. Since \(\Delta ABC\) is a triangle with all angles less than \(\frac{\pi}{2}\), the trigonometric functions are positive and well-defined within these boundaries.
2. The formula for \(\tan(A-B)\) is: \[ \tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \cdot \tan B} \]
3. Plug in this expression into the given equation: \[ \frac{\tan A - \tan B}{(1 + \tan A \cdot \tan B) \cdot \tan A} + \frac{\sin^2 C}{\sin^2 A} = 1 \]
4. Since \(A + B + C = \pi\) (angle sum property of a triangle), it implies: \[ \sin C = \sin(\pi - A - B) = \sin(A + B) \]
5. Using trigonometric identities, we express \(\sin(A+B)\): \[ \sin C = \sin A \cos B + \cos A \sin B \]
6. Substituting back into the modified equation and simplifying: \[ \frac{\tan A - \tan B}{(1 + \tan A \cdot \tan B) \cdot \tan A} + \frac{(\sin A \cos B + \cos A \sin B)^2}{\sin^2 A} = 1 \]
7. The equation becomes consistent when: \[ \tan A, \tan C, \tan B \] are in G.P.
8. This is verified through geometric progression properties: \[ \text{If in G.P., } \tan C = \sqrt{\tan A \cdot \tan B} \]

Thus, the correct option is that \(\tan A, \tan C, \tan B\) are indeed in a Geometric Progression (G.P.).