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JEE Main · 2026 · 04 Apr 2026 · Shift 1 · Trigonometric Ratios and Identities

If

A = \frac{s i n 3 ^{\circ}}{c o s 9 ^{\circ}} + \frac{s i n 9 ^{\circ}}{c o s 2 7 ^{\circ}} + \frac{s i n 2 7 ^{\circ}}{c o s 8 1 ^{\circ}}

and

B = tan 8 1^{\circ} - tan 3^{\circ}

, then

\frac{B}{A}

is equal to ____.

Answer: 2

Solution

For the general term,

\frac{sin θ}{cos 3 θ} = \frac{2 sin θ cos θ}{2 cos 3 θ cos θ} = \frac{sin 2 θ}{2 cos 3 θ cos θ} = \frac{sin ( 3 θ - θ )}{2 cos 3 θ cos θ} = \frac{1}{2} (tan 3 θ - tan θ)

. With

θ = 3^{\circ}, 9^{\circ}, 2 7^{\circ}

this telescopes:

A = \frac{1}{2} [(tan 9^{\circ} - tan 3^{\circ}) + (tan 2 7^{\circ} - tan 9^{\circ}) + (tan 8 1^{\circ} - tan 2 7^{\circ})] = \frac{1}{2} (tan 8 1^{\circ} - tan 3^{\circ}) = \frac{B}{2}

. Therefore

\frac{B}{A} = 2

.

JEE Main · 2026 · 04 Apr 2026 · Shift 1 · Integer answer

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