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JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Trigonometric Ratios and Identities

If

\frac{c o s ^{2} 4 8 ^{\circ} - s i n ^{2} 1 2 ^{\circ}}{s i n ^{2} 2 4 ^{\circ} - s i n ^{2} 6 ^{\circ}} = \frac{α + β 5}{2}

, where

α, β \in N

, then

α + β

is equal to

Answer: 4

Solution

Using

cos^{2} A - sin^{2} B = cos (A + B) cos (A - B)

and

sin^{2} A - sin^{2} B = sin (A + B) sin (A - B)

, the numerator is

cos 6 0^{\circ} cos 3 6^{\circ}

and the denominator is

sin 3 0^{\circ} sin 1 8^{\circ}

. Hence the expression is

\frac{cos 3 6 ^{\circ}}{sin 1 8 ^{\circ}} = \frac{( 5 + 1 ) /4}{( 5 - 1 ) /4} = \frac{5 + 1}{5 - 1} = \frac{( 5 + 1 ) ^{2}}{4} = \frac{3 + 5}{2} .

Therefore

α = 3

,

β = 1

and

α + β = 4

.

JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Integer answer

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