Question

If ax + b (sec (tan^–1 x)) = c and ay + b (sec.(tan^–1 y)) = c, then `(x + y)/(1 - xy)` = ______.

Options

• `(ac)/(a^2 + c^2)`

• `(2ac)/(a - c)`

• `(2ac)/(a^2 - c^2)`

• `(a + c)/(1 - ac)`

Answer 1

If ax + b (sec (tan^–1 x)) = c and ay + b (sec.(tan^–1 y)) = c, then `(x + y)/(1 - xy)` = `underlinebb((2ac)/(a^2 - c^2))`.

Explanation:

Let tan^–1 x = α and tan^–1 y = β

`\implies` tan α = x, tan β = y

The given system of equations is

a tan α + b sec α = c and a tan β + b sec β = c

∴ α and β are the roots of a tan θ + b sec θ = c

`\implies` (b sec θ)² = (c – a tan θ)²

`\implies` (a² – b²) tan² θ – 2ac tan θ + c² – b² = 0

`\implies` tan α + tan β = `(2ac)/(a^2 - b^2)` and

tan α tan β = `(c^2 - b^2)/(a^2 - b^2)`

`\implies` x + y = `(2ac)/(a^2 - b^2)` and xy = `(c^2 - b^2)/(a^2 - b^2)`

`\implies` 1 – xy = `(a^2 - c^2)/(a^2 - b^2)`

`\implies` `(x + y)/(1 - xy) = (2ac)/(a^2 - c^2)`