Если α-β=π/2, то А=sinα-sinβ/cosα-cosβ=?

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Дано, что α - β = π/2.

Так как sin(π/2 - x) = cos(x) и cos(π/2 - x) = sin(x), мы можем переписать A:

A = (sin(α) - sin(β))/(cos(α) - cos(β))

A = (sin(α) - sin(β))/(cos(α) - cos(β))*(sin(α) + sin(β))/(sin(α) + sin(β))

A = [(sin(α)^2 - sin(β)^2) + sin(α)sin(β) - sin(α)sin(β)] / [cos(α)sin(α) + sin(β)cos(α) - sin(α)cos(β) - cos(α)sin(β)]

A = [sin(α)^2 - sin(β)^2] / [cos(α)sin(α) - sin(α)cos(β)]

A = [sin(α)^2 - sin(β)^2] / sin(α)(cos(α) - cos(β))

A = (sin(α) + sin(β))(sin(α) - sin(β)) / sin(α)(cos(α) - cos(β))

A = (cos(α) + cos(β))(sin(α) - sin(β)) / sin(α)(cos(α) - cos(β))

Подставляем sin(α) = cos(β) и sin(β) = cos(α):

A = (cos(α) + sin(α))(cos(α) - sin(α)) / sin(α)(cos(α) - sin(α))

A = (cos^2(α) - sin^2(α)) / sin(α)cos(α)

A = cos(2α) / sin(2α)

A = cot(2α)

Таким образом, если α - β = π/2, то A = cot(2α).