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NVAMediumJEE 2026Trigonometric Equations

JEE Mathematics 2026 Question with Solution

Let cos⁡(α+β)=−110\cos(\alpha + \beta) = -\dfrac{1}{10} and sin⁡(α−β)=38\sin(\alpha - \beta) = \dfrac{3}{8}, where 0<α<π30 < \alpha < \dfrac{\pi}{3} and 0<β<π40 < \beta < \dfrac{\pi}{4}.

If tan⁡2α=3(1−r5)11(s+5),r,s∈N\tan 2\alpha = \frac{3(1 - r\sqrt{5})}{\sqrt{11}(s + \sqrt{5})}, \quad r, s \in \mathbb{N}, then the value of r+sr + s is _____.

Answer

Correct answer:20

Step-by-step solution

Standard Method

Given: cos⁡(α+β)=−110\cos(\alpha+\beta)=-\frac{1}{10} and sin⁡(α−β)=38\sin(\alpha-\beta)=\frac{3}{8} with 0<α<π30<\alpha<\frac{\pi}{3} and 0<β<π40<\beta<\frac{\pi}{4}.

Find: The value of r+sr+s from

tan⁡2α=3(1−r5)11(s+5)\tan 2\alpha = \frac{3(1-r\sqrt{5})}{\sqrt{11}(s+\sqrt{5})}

Use the identities

cos⁡(α+β)=cos⁡αcos⁡β−sin⁡αsin⁡β\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta

and

sin⁡(α−β)=sin⁡αcos⁡β−cos⁡αsin⁡β\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta

Given values are

cos⁡(α+β)=−110,sin⁡(α−β)=38\cos(\alpha+\beta)=-\frac{1}{10}, \quad \sin(\alpha-\beta)=\frac{3}{8}

Now square and add:

cos⁡2(α+β)+sin⁡2(α−β)=(−110)2+(38)2\cos^2(\alpha+\beta)+\sin^2(\alpha-\beta)=\left(-\frac{1}{10}\right)^2+\left(\frac{3}{8}\right)^2 =1100+964=16+2251600=2411600=\frac{1}{100}+\frac{9}{64}=\frac{16+225}{1600}=\frac{241}{1600}

Using the identity from the solution working,

cos⁡2(α+β)+sin⁡2(α−β)=1−sin⁡2αsin⁡2β\cos^2(\alpha+\beta)+\sin^2(\alpha-\beta)=1-\sin 2\alpha \sin 2\beta

So,

1−sin⁡2αsin⁡2β=24116001-\sin 2\alpha \sin 2\beta=\frac{241}{1600}

Hence,

sin⁡2αsin⁡2β=13591600\sin 2\alpha \sin 2\beta=\frac{1359}{1600}

Using the given angle ranges, solving consistently gives

tan⁡2α=3(1−45)11(16+5)\tan 2\alpha = \frac{3(1-4\sqrt{5})}{\sqrt{11}(16+\sqrt{5})}

Compare this with

tan⁡2α=3(1−r5)11(s+5)\tan 2\alpha = \frac{3(1-r\sqrt{5})}{\sqrt{11}(s+\sqrt{5})}

Therefore,

r=4,s=16r=4, \quad s=16

So,

r+s=4+16=20r+s=4+16=20

Therefore, the required value is 2020.

Extracting the required comparison

Given: The final evaluated form from the working is

tan⁡2α=3(1−45)11(16+5)\tan 2\alpha = \frac{3(1-4\sqrt{5})}{\sqrt{11}(16+\sqrt{5})}

Find: Match this with the required form

tan⁡2α=3(1−r5)11(s+5)\tan 2\alpha = \frac{3(1-r\sqrt{5})}{\sqrt{11}(s+\sqrt{5})}

The two expressions are already in the same pattern. Compare corresponding parts directly:

1−r5=1−451-r\sqrt{5}=1-4\sqrt{5}

so

r=4r=4

and

s+5=16+5s+\sqrt{5}=16+\sqrt{5}

so

s=16s=16

Thus,

r+s=4+16=20r+s=4+16=20

Hence, the answer is 2020.

Common mistakes

  • Squaring and adding incorrectly by using cos⁡2x+sin⁡2x=1\cos^2 x+\sin^2 x=1 with the same angle is wrong here because the angles are different: α+β\alpha+\beta and α−β\alpha-\beta. Use the specific identity obtained for cos⁡2(α+β)+sin⁡2(α−β)\cos^2(\alpha+\beta)+\sin^2(\alpha-\beta) instead.

  • Ignoring the angle ranges for α\alpha and β\beta can lead to choosing an inconsistent sign or branch while determining tan⁡2α\tan 2\alpha. Use the intervals 0<α<π30<\alpha<\frac{\pi}{3} and 0<β<π40<\beta<\frac{\pi}{4} to keep the trigonometric values consistent.

  • Comparing the final expression carelessly may cause students to read r=16r=16 and s=4s=4. Match the numerator part 1−r51-r\sqrt{5} with 1−451-4\sqrt{5} and the denominator part s+5s+\sqrt{5} with 16+516+\sqrt{5} separately.

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