Trigonometry & Inverse Trigonometry
Condition on tan A, tan B, tan C — Progression Type
nta_pyq_2026_jan
Grade 11
Question:
If $\dfrac{\tan(A-B)}{\tan A}+\dfrac{\sin^2C}{\sin^2A}=1$, $A,B,C\in\left(0,\dfrac{\pi}{2}\right)$, then
tan A, tan C, tan B are in A.P.
tan A, tan B, tan C are in G.P.
tan A, tan C, tan B are in G.P.
tan A, tan B, tan C are in A.P.
Step-by-Step Solution
Key Concept: Rearrange: $\tfrac{\tan(A-B)}{\tan A}=1-\tfrac{\sin^2C}{\sin^2A}=\tfrac{\sin^2A-\sin^2C}{\sin^2A}$. This simplifies to $\tan(A-B)\tan A=\sin^2A-\sin^2C$.
$\tan A,\tan C,\tan B$ are in G.P.
Correct Answer: 3