(c,d)

If – 1 ≤ x < 0, then – ≤ sin –1 x < 0. Also 0 < 2 cot –1 (y 2 – 2y) < 2 π

∴ – < sin –1 x + 2 cot –1 (y 2 – 2y) < 2 π

∴ there is no solution in this case.

thus x can not be negative ........(i)

Now if x ≥ 0, then 0 ≤ sin –1 x ≤ ⇒ ≤ cot –1 (y 2 – 2y) < π

⇒ y 2 – 2y ≤ – 1 ⇒ y = 1

since for y = 1, we have 2 cot –1 (y 2 – 2y) = 2 cot –1 (–1) =

∴ sin –1 x = i.e. x = 1 ∴ the solution is x = 1, y = 1

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