All possible values of θ ∈ [0, 2π] for which sin 2θ + tan 2θ > 0 lie in :

Solution

Maths sir · 4 steps

Factor out tan 2θ and the leftover is 2 cos² θ, which can never be negative.

1. Factor the expression. Write sin 2θ = tan 2θ cos 2θ. Then the left side becomes tan 2θ (1 + cos 2θ), and 1 + cos 2θ = 2 cos² θ.

   sin 2θ + tan 2θ = 2 cos² θ · tan 2θ > 0

2. Reduce the condition. 2 cos² θ ≥ 0 always, and it is 0 only at θ = π/2, 3π/2. So the inequality holds exactly when tan 2θ > 0 and cos θ ≠ 0.
3. Sign of tangent. For θ ∈ [0, 2π], 2θ ∈ [0, 4π]. tan 2θ > 0 when 2θ is in the first or third quadrant: (0, π/2) ∪ (π, 3π/2) ∪ (2π, 5π/2) ∪ (3π, 7π/2). Halve each interval to get θ.
4. So. θ ∈ (0, π/4) ∪ (π/2, 3π/4) ∪ (π, 5π/4) ∪ (3π/2, 7π/4). The open intervals already leave out π/2 and 3π/2, so the answer is (d).

The idea to keep. When a trig inequality mixes sin and tan of the same angle, factor out the common function. A leftover square like cos² θ only removes isolated points; it never flips the sign.

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