Trigonometric RatioseasyPYQ · JEE Main · 4 Apr 2026 · Shift 1 (Morning)Free

Telescoping Trig Sum: B/A = 2 | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question
If A=sin⁡3∘cos⁡9∘+sin⁡9∘cos⁡27∘+sin⁡27∘cos⁡81∘A=\dfrac{\sin3^\circ}{\cos9^\circ}+\dfrac{\sin9^\circ}{\cos27^\circ}+\dfrac{\sin27^\circ}{\cos81^\circ} and B=tan⁡81∘−tan⁡3∘B=\tan81^\circ-\tan3^\circ, then BA\dfrac{B}{A} is equal to
Solution
Answer: 2 (± 0.01)
Step 1: Consider the general term E=sin⁡θcos⁡3θE=\dfrac{\sin\theta}{\cos3\theta}. Multiply numerator and denominator by 2cos⁡θ2\cos\theta:
E=2sin⁡θcos⁡θ2cos⁡3θcos⁡θ=sin⁡2θ2cos⁡3θcos⁡θ=sin⁡(3θ−θ)2cos⁡3θcos⁡θ=12(tan⁡3θ−tan⁡θ).E=\frac{2\sin\theta\cos\theta}{2\cos3\theta\cos\theta}=\frac{\sin2\theta}{2\cos3\theta\cos\theta}=\frac{\sin(3\theta-\theta)}{2\cos3\theta\cos\theta}=\frac12\big(\tan3\theta-\tan\theta\big).
 Step 2: Apply with θ=3∘,9∘,27∘\theta=3^\circ,9^\circ,27^\circ:
A=12[(tan⁡9∘−tan⁡3∘)+(tan⁡27∘−tan⁡9∘)+(tan⁡81∘−tan⁡27∘)]=12(tan⁡81∘−tan⁡3∘).A=\frac12\big[(\tan9^\circ-\tan3^\circ)+(\tan27^\circ-\tan9^\circ)+(\tan81^\circ-\tan27^\circ)\big]=\frac12(\tan81^\circ-\tan3^\circ).
 Step 3: So A=12BA=\dfrac12 B, giving
BA=2.\frac{B}{A}=2.
 Correct answer: 2
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