If $\frac{\tan (\mathrm{A}-\mathrm{B})}{\tan \mathrm{A}}+\frac{\sin ^{2} \mathrm{C}}{\sin ^{2} \mathrm{~A}}=1, \mathrm{~A}, \mathrm{~B}, \mathrm{C} \in\left(0, \frac{\pi}{2}\right)$, then

(A) $\tan \mathrm{A}, \tan \mathrm{C}, \tan \mathrm{B}$ are in G.P.

(B) $\tan \mathrm{A}, \tan \mathrm{B}, \tan \mathrm{C}$ are in G.P.

(C) $\tan A, \tan C, \tan B$ are in A.P.

(D) $\tan \mathrm{A}, \tan \mathrm{B}, \tan \mathrm{C}$ are in A.P.

JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Properties of Triangle

\frac{t a n ( A - B )}{t a n A} + \frac{s i n ^{2} C}{s i n ^{2} A} = 1, A, B, C \in (0, \frac{π}{2})

, then

Answer: (A)

Writing

tan (A - B) = \frac{t a n A - t a n B}{1 + t a n A t a n B}

, the first term gives

1 - \frac{t a n ( A - B )}{t a n A} = \frac{t a n B ( 1 + t a n ^{2} A )}{t a n A ( 1 + t a n A t a n B )}

. Also

\frac{s i n ^{2} C}{s i n ^{2} A} = \frac{t a n ^{2} C ( 1 + t a n ^{2} A )}{t a n ^{2} A ( 1 + t a n ^{2} C )}

. Equating these and cancelling

(1 + tan^{2} A)

gives

\frac{t a n B}{t a n A ( 1 + t a n A t a n B )} = \frac{t a n ^{2} C}{t a n ^{2} A ( 1 + t a n ^{2} C )}

, which reduces to

tan^{2} C = tan A tan B

. Thus

tan A, tan C, tan B

are in geometric progression.

JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Single correct option