Let S = { θ ∈ [ - π ,π ] - { ± π/2 }:sin θ tan θ + tan θ = sin 2θ }.
If T = Σ_(θ ∈ S) cos 2θ , then T + n(S) is equal to :

Solution

Maths sir · 4 steps

Turn everything into sine and cosine, then pull out the common factor sin θ instead of cancelling it.

1. Factor it. Write tan θ = (sin θ)/(cos θ) and sin 2θ = 2 sin θ cos θ. Multiply through by cos θ, which is allowed because θ ≠ ±π/2. Then take sin θ out as a common factor and use 2 cos² θ = 2 - 2 sin² θ.

   sin θ(sin θ + 1 - 2 cos² θ) = 0 ⇒ sin θ (2 sin θ - 1)(sin θ + 1) = 0

2. Collect the roots. sin θ = 0 gives θ = -π, 0, π. sin θ = 1/2 gives θ = π/6, 5π/6. sin θ = -1 gives only θ = -π/2, which is excluded from the set, so n(S) = 5.
3. Find T. Use cos 2θ = 1 - 2 sin² θ. Each of the three roots with sin θ = 0 gives 1, and each of the two roots with sin θ = 1/2 gives 1/2. So T = 3 + 1 = 4.
4. So. T + n(S) = 4 + 5 = 9, which is option (b).

The idea to keep. Never cancel a factor like sin θ from both sides. Factor it out, or you lose roots. Then check every root against the excluded values.

Answer from the NTA final answer key. Solution by PrepWiser.