JEE Main 2016 · 10 Apr · Shift 1 · Mathematics Q61

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MathematicsSingle correct+4 / -1Question 61 of 90

Let $P=\{\theta:\sin\theta-\cos\theta=\sqrt2\cos\theta\}$ and $Q=\{\theta:\sin\theta+\cos\theta=\sqrt2\sin\theta\}$ be two sets. Then :

P\subset Q

and

Q-P\ne\phi

Q\not\subset P

P\not\subset Q

P = Q

Answer and worked solution

Official answer: D

P: $\sin\theta=(1+\sqrt2)\cos\theta$ , i.e. $\tan\theta=\sqrt2+1$ . Q: $\cos\theta=(\sqrt2-1)\sin\theta$ , i.e. $\tan\theta=\frac1{\sqrt2-1}=\sqrt2+1$ . The two conditions are the same, so P = Q.

Topics:Trigonometric equations Sets and operations

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JEE Main 2016 · 10 Apr · Shift 1, Mathematics question 61