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JEE Main 202628 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
28 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

JEE Main 2026 · 28 January · Morning Shift · Mathematics · Trigonometric Ratios & Identities

If (A-B) A + ^ 2 C ^ 2 ~A =1, ~A, ~B, C ∈(0, (π)/(2)) , then

If tan⁡(A−B)tan⁡A+sin⁡2Csin⁡2 A=1, A, B,C∈(0,π2)\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

Options

  1. Atan⁡A,tan⁡C,tan⁡B\tan \mathrm{A}, \tan \mathrm{C}, \tan \mathrm{B} are in A.P.
  2. Btan⁡A,tan⁡B,tan⁡C\tan A, \tan B, \tan C are in G.P.
  3. Ctan⁡A,tan⁡C,tan⁡B\tan A, \tan C, \tan B are in G.P.
  4. Dtan⁡A,tan⁡B,tan⁡C\tan \mathrm{A}, \tan \mathrm{B}, \tan \mathrm{C} are in A.P.

Correct answer

C. tan⁡A,tan⁡C,tan⁡B\tan A, \tan C, \tan B are in G.P.

Step-by-step solution

Starting with tan⁡(A−B)tan⁡A+sin⁡2Csin⁡2A=1\frac{\tan(A-B)}{\tan A} + \frac{\sin^2 C}{\sin^2 A} = 1, rearrange to tan⁡(A−B)tan⁡A=1−sin⁡2Csin⁡2A\frac{\tan(A-B)}{\tan A} = 1 - \frac{\sin^2 C}{\sin^2 A}.
⇒tan⁡(A−B)tan⁡A=sin⁡2A−sin⁡2C\Rightarrow \tan(A-B) \tan A = \sin^2 A - \sin^2 C.
Applying the identity sin⁡2θ=tan⁡2θ1+tan⁡2θ\sin^2 \theta = \frac{\tan^2 \theta}{1+\tan^2 \theta} and simplifying leads to a relationship between the tangent values.
The constraint reduces to (tan⁡C)2=tan⁡A⋅tan⁡B(\tan C)^2 = \tan A \cdot \tan B, which means tan⁡A\tan A, tan⁡C\tan C, and tan⁡B\tan B form a geometric progression with common ratio r=tan⁡Ctan⁡Ar = \frac{\tan C}{\tan A}.
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