Verified Solution • 2025–26 Session moderate

QJEEM22Ma286b: JEE Main 2022 Mathematics - Class 11 JEEM22Ma286b Solution: Let . If , then T + n(S) is equal to :

Let S={θ∈[−π,π]−{±  π2}:sin⁡θtan⁡θ+tan⁡θ=sin⁡2θ}S = \left\{ {\theta \in [ - \pi ,\pi ] - \left\{ { \pm \,\,{\pi \over 2}} \right\}:\sin \theta \tan \theta + \tan \theta = \sin 2\theta } \right\} .

If T=∑θ ∈ Scos⁡2θT = \sum\limits_{\theta \, \in \,S}^{} {\cos 2\theta } , then T + n(S) is equal to :

Multiple Choice Options

  • A 7 + 3\sqrt 3
  • B 9
  • C 8 + 3\sqrt 3
  • D 10

Step-by-Step Verified Solution

tan⁡θ(sin⁡θ+1)−sin⁡2θ=0\tan \theta(\sin \theta+1)-\sin 2 \theta=0

amp;tan⁡θ(sin⁡θ+1−2cos⁡2θ)=0amp;⇒tan⁡θ=0 or 2sin⁡2θ+sin⁡θ−1=0amp;⇒(2sin⁡θ+1)(sin⁡θ−1)=0amp;⇒sin⁡θ=−12 or 1\begin{aligned}&\tan \theta\left(\sin \theta+1-2 \cos ^{2} \theta\right)=0 \\\\&\Rightarrow \tan \theta=0 \text { or } 2 \sin ^{2} \theta+\sin \theta-1=0 \\\\&\Rightarrow(2 \sin \theta+1)(\sin \theta-1)=0 \\\\&\Rightarrow \sin \theta=\frac{-1}{2} \text { or } 1\end{aligned}

But, sin⁡θ=1\sin \theta=1 not possible

θ=0,π,−π,−π6,−5π6\theta=0, \pi,-\pi,-\frac{\pi}{6}, \frac{-5 \pi}{6}

n(S)=5\mathrm{n}(\mathrm{S})=5

T=∑cos⁡2θ=cos⁡0∘+cos⁡2π+cos⁡(−2π)+cos⁡(−5π3)+cos⁡(−π3)T=\sum \cos 2 \theta=\cos 0^{\circ}+\cos 2 \pi+\cos (-2 \pi)+\cos \left(-\frac{5 \pi}{3}\right)+\cos \left(-\frac{\pi}{3}\right)

= 4

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