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Mathematics · Trigonometry Ratios and Identities

JEE Main 2026 — 22 January, Morning Shift — Question 23

If cos⁡248∘−sin⁡212∘sin⁡224∘−sin⁡26∘=α+β52\frac{\cos ^{2} 48^{\circ}-\sin ^{2} 12^{\circ}}{\sin ^{2} 24^{\circ}-\sin ^{2} 6^{\circ}}=\frac{\alpha+\beta \sqrt{5}}{2}, where α,β∈N\alpha, \beta \in \mathbb{N}, then α+β\alpha+\beta is equal to ____\_\_\_\_

Answer: 4

Numerical answer — enter this value.

Step-by-step solution

Use sin⁡(A+B)sin⁡(A+B)=sin⁡2 A−sin⁡2 B\sin (\mathrm{A}+\mathrm{B}) \sin (\mathrm{A}+\mathrm{B})=\sin ^{2} \mathrm{~A}-\sin ^{2} \mathrm{~B}

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

cos⁡60∘cos⁡36∘sin⁡30∘sin⁡18∘=5+15−1×5+15+1=(5+1)24\frac{\cos 60^{\circ} \cos 36^{\circ}}{\sin 30^{\circ} \sin 18^{\circ}}=\frac{\sqrt{5}+1}{\sqrt{5}-1} \times \frac{\sqrt{5}+1}{\sqrt{5}+1}=\frac{(\sqrt{5}+1)^{2}}{4}

=3+52=\frac{3+\sqrt{5}}{2}

α=3;β=1\alpha=3 ; \beta=1 So, (α+β)=4(\alpha+\beta)=4

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
22 January, Morning Shift
Subject
Mathematics
Chapter
Trigonometry Ratios and Identities
Topic
Trigonometric Ratios of Allied Angles