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← JEE Main 2022 · 24 Jun · Shift 1
MathematicsSingle correct+4 / -1Question 19 of 90

JEE Main 2022 · 24 Jun · Shift 1, Mathematics question 19

Let S={θ∈[−π,π]−{±π2}:sin⁡θtan⁡θ+tan⁡θ=sin⁡2θ}S=\left\{\theta\in[-\pi,\pi]-\left\{\pm\dfrac\pi2\right\}:\sin\theta\tan\theta+\tan\theta=\sin2\theta\right\}. If T=∑θ∈Scos⁡2θT=\displaystyle\sum_{\theta\in S}\cos2\theta, then T+n(S)T+n(S) is equal to :

Choose one option
Answer and worked solution

Official answer: B

Factor. tan⁡θ(sin⁡θ+1)=2sin⁡θcos⁡θ\tan\theta(\sin\theta+1)=2\sin\theta\cos\theta means sin⁡θ[1+sin⁡θcos⁡θ−2cos⁡θ]=0\sin\theta\left[\dfrac{1+\sin\theta}{\cos\theta}-2\cos\theta\right]=0.

Case sin⁡θ=0\sin\theta=0: θ=−π,0,π\theta=-\pi,0,\pi.

Other case: 1+sin⁡θ=2(1−sin⁡2θ)1+\sin\theta=2(1-\sin^2\theta), so 2sin⁡2θ+sin⁡θ−1=02\sin^2\theta+\sin\theta-1=0: sin⁡θ=12\sin\theta=\frac12 (sin⁡θ=−1\sin\theta=-1 is excluded). So θ=π6,5π6\theta=\frac\pi6,\frac{5\pi}6.

Count and sum. n(S)=5n(S)=5 and T=1+1+1+cos⁡π3+cos⁡5π3=4T=1+1+1+\cos\frac\pi3+\cos\frac{5\pi}3=4. So T+n(S)=9T+n(S)=9.

(B)

Topics:Trigonometric equations
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