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Let S be the solution set of sinθ tanθ + tanθ = sin2θ.Such that θ ∈ (-π, π), θ ≠ ± \(\frac\pi2\) and t = \(\sum\)cos 2θ, where θ is solution of above equation then value of t + n (S) is,(1) 8(2) 5(3) 6(4) 9 |
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Answer» Correct option is (2) 5 sinθ tanθ + tanθ = sin2θ tanθ (sinθ + 1) - 2 sinθ cosθ = 0 \(sin \theta \left(\frac{(sin\theta + 1)}{cos\theta} - 2 cos\theta\right) = 0\) sinθ = 0 θ = 0 or sinθ + 1 - 2 cos2θ = 0 sinθ + 1 - 2 (1 - sin2θ) = 0 2sin2θ + sinθ - 1 = 0 sinθ = -1, \(\frac12\) \(\theta = -\frac\pi2,\frac\pi6,\frac{5\pi}6\) Hence, \(S = \{0, \frac\pi6,\frac{5\pi}6\}\) ⇒ n(5) = 3 \(t = \sum (\theta(2\theta)) = cos(0) + cos(\frac\pi3) + cos(\frac{5\pi}3)\) \(= 1 +\frac12 + \frac12 = 2\) t + n(5) = 2 + 3 = 5 |
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