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If cos⁻¹ ((1-x²)/(1+x²))+cos⁻¹ ((1-y²)/(1+y²))=(π)/(2), with xy<1, which relation must hold?

1. x - y - xy = 1
2. x - y + xy = 1
3. x - y - xy = 1
4. x + y + xy = 1

Correct answer: x + y + xy = 1

Solution

Since arccos((1-x^2)/(1+x^2))=2 arctan x, the equation becomes 2 arctan x + 2 arctan y = pi/2, so arctan x + arctan y = pi/4. Then (x+y)/(1-xy)=1, giving x+y+xy=1.

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