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Mathematics · Trigonometry Ratios and Identities

JEE Main 2026 — 4 April, Morning Shift — Question 45

If A=sin⁡3∘cos⁡9∘+sin⁡9∘cos⁡27∘+sin⁡27∘cos⁡81∘\mathrm{A} = \frac{\sin 3^{\circ}}{\cos 9^{\circ}} +\frac{\sin 9^{\circ}}{\cos 27^{\circ}} +\frac{\sin 27^{\circ}}{\cos 81^{\circ}} and B=tan⁡81∘−tan⁡3∘\mathrm{B} = \tan 81^{\circ} - \tan 3^{\circ}, then BA\frac{\mathrm{B}}{\mathrm{A}} is equal to ____.

Answer: 2

Numerical answer — enter this value.

Step-by-step solution

Consider E=sin⁡θcos⁡3θ⇒E=2sin⁡θcos⁡θ2cos⁡3θcos⁡θE=\frac{\sin \theta}{\cos 3 \theta} \Rightarrow E=\frac{2 \sin \theta \cos \theta}{2 \cos 3 \theta \cos \theta} ⇒E=sin⁡2θ2cos⁡3θcos⁡θ⇒E=sin⁡(30−θ)2cos⁡3θcos⁡θ\Rightarrow \mathrm{E}=\frac{\sin 2 \theta}{2 \cos 3 \theta \cos \theta} \Rightarrow \mathrm{E}=\frac{\sin (30-\theta)}{2 \cos 3 \theta \cos \theta} ⇒E=12[tan⁡3θ−tan⁡θ]\Rightarrow \mathrm{E}=\frac{1}{2}[\tan 3 \theta-\tan \theta] A=12[tan⁡9∘−tan⁡3∘+tan⁡27∘−tan⁡9∘+tan⁡81∘−tan⁡27∘]\mathrm{A}=\frac{1}{2}\left[\tan 9^{\circ}-\tan 3^{\circ}+\tan 27^{\circ}-\tan 9^{\circ}+\tan 81^{\circ}-\tan 27^{\circ}\right] ∴A=12[tan⁡81∘−tan⁡3∘]\therefore \mathrm{A}=\frac{1}{2}\left[\tan 81^{\circ}-\tan 3^{\circ}\right] ∴BA=2\therefore \frac{\mathrm{B}}{\mathrm{A}}=2

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
4 April, Morning Shift
Subject
Mathematics
Chapter
Trigonometry Ratios and Identities
Topic
Introduction to Trigonometry
If A = frac sin 3 ° cos 9 ° +frac sin 9 ° cos 27 ° +frac sin 27 ° cos… | JEE Main 2026 PYQ with Solution · DhiX AI