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If $(x, y)$ satisfy $tan − 1 1 − y 1 + y + sin − 1 x − 1 + cos − 1 xy 2 = 12 cos − 1 x (1 − x)$ , then

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x + 2 y = 3 + 1

2 x + y = 3 + 1

x + 3 y = 4

3 x + y = 4

answer is A.

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Detailed Solution

∴ X = 1 (domain) ⇒ tan − 1 1 − y 1 + y + cos − 1 | y | = π 4

y > 0 : tan − 1 1 − y 1 + y + cos − 1 y = π 4

Let

cos − 1 y = θ; θ ∈ 0, π 2

tan − 1 tan θ 2 + θ = π 4

θ = π 6 ⇒ y = 32

y < 0 : tan − 1 1 − y 1 + y + π − cos − 1 y = π 4

Let

cos − 1 y = θ; θ ∈ π 2, π

⇒ tan − 1 tan θ 2 + π − θ = π 4 ⇒ θ = 3 π 2 (not possible)

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