Let S be the solution set of sinθ tanθ + tanθ = sin^2θ. Such that θ ∈ (-π, π), θ ≠ ± π/2 and t = cos 2θ,

Question

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

Answer 1

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