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JEE Main · 2022 · 24 Jun 2022 · Shift 1 · Trigonometric Equations

Let

S = {θ \in [- π, π] - {\pm \frac{π}{2}} : sin θ tan θ + tan θ = sin 2 θ}

.
If

T = \sum_{θ \in S} cos 2 θ

, then

T + n (S)

is equal

(A) $7 + 3$
(B)9✓ Correct
(C) $8 + 3$
(D)10

Answer: (B)

Solution

From

sin θ tan θ + tan θ = sin 2 θ

:

tan θ (sin θ + 1) = \frac{2 t a n θ}{1 + t a n ^{2} θ}

Either

tan θ = 0

giving

θ = - π, 0, π

(3 values), or

sin θ + 1 = 2 cos^{2} θ = 2 (1 - sin^{2} θ)

.

The latter simplifies to

2 sin^{2} θ + sin θ - 1 = 0

, giving

sin θ = \frac{1}{2}

(since

sin θ = - 1

is excluded). Thus

θ = \frac{π}{6}, \frac{5 π}{6}

.

So

n (S) = 5

. Now

T = cos (0) + cos (- 2 π) + cos (2 π) + cos (\frac{π}{3}) + cos (\frac{5 π}{3}) = 1 + 1 + 1 + \frac{1}{2} + \frac{1}{2} = 4

.

Therefore

T + n (S) = 4 + 5 = 9

.

JEE Main · 2022 · 24 Jun 2022 · Shift 1 · Single correct option

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