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← IIT-JEE 2012 · Paper 1
MathematicsMultiple correct+4Question 51 of 60

IIT-JEE 2012 · Paper 1, Mathematics question 51

Let θ,φ∈[0,2π]\theta,\varphi\in[0,2\pi] be such that
2cos⁡θ(1−sin⁡φ)=sin⁡2θ(tan⁡θ2+cot⁡θ2)cos⁡φ−1,2\cos\theta(1-\sin\varphi)=\sin^2\theta\left(\tan\frac\theta2+\cot\frac\theta2\right)\cos\varphi-1,
tan⁡(2π−θ)>0\tan(2\pi-\theta)>0 and −1<sin⁡θ<−32-1<\sin\theta<-\dfrac{\sqrt3}2. Then φ\varphi cannot satisfy

Choose one or more options
Answer and worked solution

Official answer: A, C, D

Locate θ. tan⁡(2π−θ)=−tan⁡θ>0\tan(2\pi-\theta)=-\tan\theta>0, so tan⁡θ<0\tan\theta<0; with sin⁡θ∈(−1,−32)\sin\theta\in(-1,-\frac{\sqrt3}2) this puts θ∈(3π2,5π3)\theta\in\left(\frac{3\pi}2,\frac{5\pi}3\right), where cos⁡θ∈(0,12)\cos\theta\in\left(0,\frac12\right).

Simplify. tan⁡θ2+cot⁡θ2=2sin⁡θ\tan\frac\theta2+\cot\frac\theta2=\frac2{\sin\theta}, so the right side is 2sin⁡θcos⁡φ−12\sin\theta\cos\varphi-1, and the equation becomes
2cos⁡θsin⁡φ+2sin⁡θcos⁡φ=2cos⁡θ+1  ⟹  sin⁡(θ+φ)=cos⁡θ+12∈(12,1).2\cos\theta\sin\varphi+2\sin\theta\cos\varphi=2\cos\theta+1\implies\sin(\theta+\varphi)=\cos\theta+\tfrac12\in\left(\tfrac12,1\right).

Range of φ. θ+φ∈(π6,5π6)+2kπ\theta+\varphi\in\left(\frac\pi6,\frac{5\pi}6\right)+2k\pi. With θ∈(3π2,5π3)\theta\in\left(\frac{3\pi}2,\frac{5\pi}3\right) and φ∈[0,2π]\varphi\in[0,2\pi], this gives θ+φ∈(13π6,17π6)\theta+\varphi\in\left(\frac{13\pi}6,\frac{17\pi}6\right), so
φ∈(13π6−5π3, 17π6−3π2)=(π2,4π3).\varphi\in\left(\frac{13\pi}6-\frac{5\pi}3,\ \frac{17\pi}6-\frac{3\pi}2\right)=\left(\frac\pi2,\frac{4\pi}3\right).

So φ cannot lie in the intervals of (A), (C) or (D).

(A), (C), (D)

Topics:Trigonometric equationsTrigonometric ratios and identities
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