Question

If \[ \frac{\tan(A-B)}{\tan A}+\frac{\sin^2 C}{\sin^2 A}=1, \quad A,B,C\in\left(0,\frac{\pi}{2}\right), \] then:

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

Hint:

In trigonometric problems involving \( A+B+C=\frac{\pi}{2} \), always try converting everything into tangents for faster conclusions.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We need to manipulate the trigonometric identity provided to find a relationship between the tangents of the angles.
Step 2: Detailed Explanation:
The given equation is:
\[ \frac{\sin^2 C}{\sin^2 A} = 1 - \frac{\tan(A - B)}{\tan A} = \frac{\tan A - \tan(A - B)}{\tan A} \] Using the identity \(\tan X - \tan Y = \frac{\sin(X - Y)}{\cos X \cos Y}\):
\[ \frac{\sin^2 C}{\sin^2 A} = \frac{\sin(A - (A - B))}{\cos A \cos(A - B)} \cdot \frac{\cos A}{\sin A} = \frac{\sin B}{\sin A \cos(A - B)} \] \[ \sin^2 C = \frac{\sin A \sin B}{\cos(A - B)} = \frac{\sin A \sin B}{\cos A \cos B + \sin A \sin B} \] Dividing the numerator and denominator by \(\cos A \cos B\):
\[ \sin^2 C = \frac{\tan A \tan B}{1 + \tan A \tan B} \] We also know that \(\sin^2 C = \frac{\tan^2 C}{1 + \tan^2 C}\).
Equating the two expressions:
\[ \frac{\tan^2 C}{1 + \tan^2 C} = \frac{\tan A \tan B}{1 + \tan A \tan B} \] By comparison or cross-multiplication:
\[ \tan^2 C = \tan A \tan B \] This implies that \(\tan A, \tan C, \tan B\) are in Geometric Progression.
Step 3: Final Answer:
\(\tan A, \tan C, \tan B\) are in G.P.