Trigonometric Ratios and Identities

Concept:The given equation can be simplified using trigonometric identities to find a relation between $\tan A$, $\tan B$, and $\tan C$.Explanation:Start with the given equation:$\frac{\tan(A-B)}{\tan A} + \frac{\sin^2 C}{\sin^2 A} = 1$.Rearrange to get:$\frac{\sin^2 C}{\sin^2 A} = 1 - \frac{\tan(A-B)}{\tan A} = \frac{\tan A - \tan(A-B)}{\tan A}$.Use the identity $\tan X - \tan Y = \frac{\sin(X-Y)}{\cos X \cos Y}$:$\tan A - \tan(A-B) = \frac{\sin B}{\cos A \cos(A-B)}$.Substitute into the previous expression:$\frac{\sin^2 C}{\sin^2 A} = \frac{\sin B}{\cos A \cos(A-B) \tan A}$.Since $\tan A = \frac{\sin A}{\cos A}$, simplify:$\frac{\sin^2 C}{\sin^2 A} = \frac{\sin B}{\sin A \cos(A-B)}$.Thus $\sin^2 C = \frac{\sin A \sin B}{\cos(A-B)}$.Write $\cos(A-B) = \cos A \cos B + \sin A \sin B$. Then divide numerator and denominator by $\cos A \cos B$:$\sin^2 C = \frac{\tan A \tan B}{1 + \tan A \tan B}$.But we also know $\sin^2 C = \frac{\tan^2 C}{1 + \tan^2 C}$.Equate the two expressions:$\frac{\tan^2 C}{1 + \tan^2 C} = \frac{\tan A \tan B}{1 + \tan A \tan B}$.Cross-multiply and simplify:$\tan^2 C + \tan^2 C \tan A \tan B = \tan A \tan B + \tan^2 C \tan A \tan B$$\Rightarrow \tan^2 C = \tan A \tan B$.Therefore, $\tan A$, $\tan C$, $\tan B$ are in geometric progression (G.P.).Answer:B. $\tan A , \tan C , \tan B$ are in G.P.