JEE Main 202622 Jan, Morning shiftMathematicsMediumTrigonometry118× since 2002If cos248∘−sin212∘sin224∘−sin26∘=α+β52\frac{\cos^2 48^\circ-\sin^2 12^\circ}{\sin^2 24^\circ-\sin^2 6^\circ}=\frac{\alpha+\beta\sqrt5}{2}sin224∘−sin26∘cos248∘−sin212∘=2α+β5, where α,β∈N\alpha, \beta\in\mathbb{N}α,β∈N, then α+β\alpha+\betaα+β is equal toYour answerType your answer. Integers and decimals are accepted.CheckSkip · show solution
