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NVAMediumJEE 2026Trigonometric Ratios & Identities

JEE Mathematics 2026 Question with Solution

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 the value of α+β\alpha + \beta is _____.

Answer

Correct answer:6

Step-by-step solution

Standard Method

Given:

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}

Find: α+β\alpha + \beta

Use trigonometric identities for differences of squares.

cos⁡2A−sin⁡2B=12(cos⁡2A+cos⁡2B)\cos^2 A - \sin^2 B = \frac{1}{2}(\cos 2A + \cos 2B)sin⁡2C−sin⁡2D=12(cos⁡2D−cos⁡2C)\sin^2 C - \sin^2 D = \frac{1}{2}(\cos 2D - \cos 2C)

Applying these,

cos⁡248∘−sin⁡212∘=12(cos⁡96∘+cos⁡24∘)\cos^2 48^\circ - \sin^2 12^\circ = \frac{1}{2}(\cos 96^\circ + \cos 24^\circ)sin⁡224∘−sin⁡26∘=12(cos⁡12∘−cos⁡48∘)\sin^2 24^\circ - \sin^2 6^\circ = \frac{1}{2}(\cos 12^\circ - \cos 48^\circ)

After simplification, the expression becomes

6+52\frac{6 + \sqrt{5}}{2}

Comparing with

α+β52\frac{\alpha + \beta\sqrt{5}}{2}

we get

α=6,β=1\alpha = 6, \quad \beta = 1

Therefore,

α+β=7\alpha + \beta = 7

So the required value is 77. The solution lists the final answer as 66, but the working gives α=6\alpha = 6 and β=1\beta = 1, hence α+β=7\alpha + \beta = 7.

From comparison of coefficients

Given:

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}

Find: α+β\alpha + \beta

The simplified form shown in the solution is

6+52\frac{6 + \sqrt{5}}{2}

Comparing term by term with

α+β52\frac{\alpha + \beta\sqrt{5}}{2}

gives

α=6\alpha = 6

and

β=1\beta = 1

Hence,

α+β=7\alpha + \beta = 7

Therefore, the required value is 77.

Common mistakes

  • Taking α=6\alpha = 6 directly as the final answer. This is wrong because the question asks for α+β\alpha + \beta, not only α\alpha. After comparison, add β=1\beta = 1 as well.

  • Using the identity for cos⁡2x−sin⁡2x\cos^2 x - \sin^2 x in a situation where the angles are different. Here the angles are 48∘48^\circ and 12∘12^\circ, so use the separate square identities before combining terms.

  • Missing the factor 12\frac{1}{2} while converting cos⁡2θ\cos^2\theta or sin⁡2θ\sin^2\theta into double-angle form. This changes the entire ratio and leads to an incorrect comparison with α+β52\frac{\alpha + \beta\sqrt{5}}{2}.

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