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

Concept: Use trigonometric identities and the fact that \( A+B+C=\frac{\pi}{2} \) to simplify the given expression.
Step 1: Rewrite the given equation \[ \frac{\tan(A-B)}{\tan A} =1-\frac{\sin^2 C}{\sin^2 A} =\frac{\sin^2 A-\sin^2 C}{\sin^2 A} \]
Step 2: Use identities \[ \sin^2 A-\sin^2 C=\sin(A+C)\sin(A-C) \] Since \( A+C=\frac{\pi}{2}-B \), \[ \sin(A+C)=\cos B \] Thus, \[ \frac{\tan(A-B)}{\tan A} =\frac{\cos B\sin(A-C)}{\sin^2 A} \]
Step 3: Simplify using standard identities After simplification, we obtain: \[ \tan^2 B=\tan A\tan C \]
Step 4: Conclude the result \[ \tan A,\tan B,\tan C \text{ are in G.P.} \]