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JEE Main 2026 · 4 Apr · Shift 1, Mathematics question 23

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 BA is equal to ________.

Answer and worked solution

Official answer: 2

Identity. tan⁡3x−tan⁡x=sin⁡2xcos⁡3xcos⁡x=2sin⁡xcos⁡3x\tan3x-\tan x=\dfrac{\sin2x}{\cos3x\cos x}=\dfrac{2\sin x}{\cos3x}.

Telescope. A=12[(tan⁡9∘−tan⁡3∘)+(tan⁡27∘−tan⁡9∘)+(tan⁡81∘−tan⁡27∘)]=B2A=\frac12\left[(\tan9^\circ-\tan3^\circ)+(\tan27^\circ-\tan9^\circ)+(\tan81^\circ-\tan27^\circ)\right]=\frac B2.

Ratio. BA=2\dfrac BA=2.

Answer: 2

Topics:Trigonometric ratios and identities
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