SolveItJEE Main
Mathematics · 2026

JEE Main · 22 January 2026, Shift 1 · Q24

If (cos^2 48^°- sin^2 12^°)/(sin^2 24^°- sin^2 6^°)=(α+β √ 5)/2, where α, β ∈ N, then α+β is equal to ____

If $\displaystyle \frac{\cos ^2 48^{\circ}-\sin ^2 12^{\circ}}{\sin ^2 24^{\circ}-\sin ^2 6^{\circ}}=\frac{\alpha+\beta \sqrt{5}}{2}$, where $\displaystyle \alpha, \beta \in \mathbb{N}$, then $\displaystyle \alpha+\beta$ is equal to $\displaystyle \_\_\_\_$
ShareWhatsAppTelegram

More from Trigonometric Identities

JEE Main 2026 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.