JEE Main 2026 Jan 22 Shift 1, Mathematics Q24: Trigonometric Ratios And Identities

Apply two standard identities:

${\cos }^{2}A-{\sin }^{2}B=\cos (A+B)\cos (A-B)$, so ${\cos }^{2}48deg;-{\sin }^{2}12deg;=\cos 60deg;\cos 36deg;$.

${\sin }^{2}A-{\sin }^{2}B=\sin (A+B)\sin (A-B)$, so ${\sin }^{2}24deg;-{\sin }^{2}6deg;=\sin 30deg;\sin 18deg;$.

Since $\cos 60deg;=\sin 30deg;=1/2$, the ratio simplifies to $\frac{\cos 36deg;}{\sin 18deg;}$.

Standard values: $\cos 36deg;=\frac{\sqrt{5}+1}{4}$, $\sin 18deg;=\frac{\sqrt{5}-1}{4}$.

$\frac{\cos 36deg;}{\sin 18deg;}=\frac{\sqrt{5}+1}{\sqrt{5}-1}=\frac{(\sqrt{5}+1{)}^2}{4}=\frac{6+2\sqrt{5}}{4}=\frac{3+\sqrt{5}}{2}$.

Compare with $\frac{\alpha +\beta \sqrt{5}}{2}$: $\alpha =3,\beta =1$, so $\alpha +\beta =4$.