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

Step 1: Understanding the Concept:
Use the relation \( A + B + C = \pi \implies \sin C = \sin(A+B) \).
Transform the given equation into a simpler trigonometric relation.
Step 2: Key Formula or Approach:
Given: \( \frac{\tan(A-B)}{\tan A} + \frac{\sin^2 C}{\sin^2 A} = 1 \implies \frac{\sin^2 C}{\sin^2 A} = 1 - \frac{\tan(A-B)}{\tan A} \).
Step 3: Detailed Explanation:
Using the identity \( \tan A - \tan(A-B) = \frac{\sin B}{\cos A \cos(A-B)} \):
\[ \frac{\sin^2(A+B)}{\sin^2 A} = \frac{\tan A - \tan(A-B)}{\tan A} = \frac{\sin B / (\cos A \cos(A-B))}{\sin A / \cos A} \] \[ \frac{\sin^2(A+B)}{\sin^2 A} = \frac{\sin B}{\sin A \cos(A-B)} \] \[ \sin^2(A+B) \cos(A-B) = \sin A \sin B \] Expanding \( \sin(A+B) \):
Simplifying this equation leads to \( \tan A \tan B = \tan^2 C \).
This indicates that \( \tan A, \tan C, \tan B \) are in Geometric Progression.
Step 4: Final Answer:
\( \tan A, \tan C, \tan B \) are in G.P.