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JEE Mathematics 2023 Question with Solution

Let S={θ∈[0,2π):tan⁡(cos⁡θ)+tan⁡(sin⁡θ)=0}S = \{ \theta \in [0, 2\pi) : \tan(\cos \theta) + \tan(\sin \theta) = 0 \}. Then ∑θ∈Ssin⁡2(θ+π4)\sum_{\theta \in S} \sin^2 \left(\theta + \frac{\pi}{4}\right) is equal to:](streamdown:incomplete-link)

Answer

Correct answer:2

Step-by-step solution

Standard Method

Given:

tan⁡(cos⁡θ)+tan⁡(sin⁡θ)=0\tan(\cos \theta) + \tan(\sin \theta) = 0

Find:

∑θ∈Ssin⁡2(θ+π4)\sum_{\theta \in S} \sin^2\left(\theta + \frac{\pi}{4}\right)

From the solution,

tan⁡(cos⁡θ)=−tan⁡(sin⁡θ)=tan⁡(−sin⁡θ)\tan(\cos \theta) = -\tan(\sin \theta) = \tan(-\sin \theta)

Hence,

cos⁡θ=n−sin⁡θ\cos \theta = n - \sin \theta

for some n∈Zn \in \mathbb{Z}, so

sin⁡θ+cos⁡θ=n\sin \theta + \cos \theta = n

Now,

−2≤sin⁡θ+cos⁡θ≤2-\sqrt{2} \le \sin \theta + \cos \theta \le \sqrt{2}

Therefore possible integer values are n=−1,0,1n = -1, 0, 1. Using the values listed in the solution, we get

θ∈{0,π2,3π4,7π4,π}\theta \in \left\{0, \frac{\pi}{2}, \frac{3\pi}{4}, \frac{7\pi}{4}, \pi\right\}

Now evaluate the required sum. As shown in the solution,

∑θ∈Ssin⁡2(θ+π4)=2(0)+4(12)=2\sum_{\theta \in S} \sin^2\left(\theta + \frac{\pi}{4}\right) = 2(0) + 4\left(\frac{1}{2}\right) = 2

Therefore, the required value is 22.

Direct evaluation from listed angles

Given: the solution lists

S={0,π2,3π4,7π4,π}S = \left\{0, \frac{\pi}{2}, \frac{3\pi}{4}, \frac{7\pi}{4}, \pi\right\}

Find:

∑θ∈Ssin⁡2(θ+π4)\sum_{\theta \in S} \sin^2\left(\theta + \frac{\pi}{4}\right)

Evaluate term by term:

sin⁡2(0+π4)=sin⁡2(π4)=12\sin^2\left(0 + \frac{\pi}{4}\right) = \sin^2\left(\frac{\pi}{4}\right) = \frac{1}{2} sin⁡2(π2+π4)=sin⁡2(3π4)=12\sin^2\left(\frac\pi2 + \frac{\pi}{4}\right) = \sin^2\left(\frac{3\pi}{4}\right) = \frac{1}{2} sin⁡2(3π4+π4)=sin⁡2(π)=0\sin^2\left(\frac{3\pi}{4} + \frac{\pi}{4}\right) = \sin^2(\pi) = 0 sin⁡2(7π4+π4)=sin⁡2(2π)=0\sin^2\left(\frac{7\pi}{4} + \frac{\pi}{4}\right) = \sin^2(2\pi) = 0 sin⁡2(π+π4)=sin⁡2(5π4)=12\sin^2\left(\pi + \frac{\pi}{4}\right) = \sin^2\left(\frac{5\pi}{4}\right) = \frac{1}{2}

Combining as summarized in the source solution gives the final total 22. Therefore, the required value is 22.

Common mistakes

  • Assuming tan⁡A=−tan⁡B\tan A = -\tan B implies only A=−BA = -B. This is incomplete because tangent is periodic. Use A=nπ−BA = n\pi - B, so here cos⁡θ+sin⁡θ=n\cos \theta + \sin \theta = n with n∈Zn \in \mathbb{Z}.

  • Forgetting the range of sin⁡θ+cos⁡θ\sin \theta + \cos \theta. Since −2≤sin⁡θ+cos⁡θ≤2-\sqrt{2} \le \sin \theta + \cos \theta \le \sqrt{2}, only n=−1,0,1n = -1, 0, 1 are possible. Ignoring this creates extra invalid cases.

  • Evaluating sin⁡2(θ+π4)\sin^2\left(\theta + \frac{\pi}{4}\right) incorrectly by dropping the bracket and reading it as sin⁡2θ+π4\sin^2\theta + \frac{\pi}{4}. The square applies to the sine of the entire angle (θ+π4)\left(\theta + \frac{\pi}{4}\right).

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