Question

Given `x = (sqrt(a^2 + b^2) + sqrt(a^2 - b^2))/(sqrt(a^2 + b^2) - sqrt(a^2 - b^2)`. Use componendo and dividendo to prove that: `b^2 = (2a^2x)/(x^2 + 1)`

If `x = (sqrt(a^2 + b^2) + sqrt(a^2 - b^2))/(sqrt(a^2 + b^2) - sqrt(a^2 - b^2)`, use componendo and dividendo, prove that `b^2 = (2a^2x)/(x^2 + 1)`

Answer 1

Given: `x/(1) = (sqrt(a^2 + b^2) + sqrt(a^2 - b^2))/(sqrt(a^2 + b^2) - sqrt(a^2 - b^2)`

Applying componendo and dividendo both sides, we have

`(x + 1)/(x - 1) = (sqrt(a^2 + b^2) + sqrt(a^2 - b^2) + sqrt(a^2 + b^2) - sqrt(a^2 - b^2))/(sqrt(a^2 + b^2) + sqrt(a^2 - b^2) - sqrt(a^2 + b^2) - sqrt(a^2 - b^2)`

⇒ `(x + 1)/(x - 1) = (2sqrt(a^2 + b^2))/(2sqrt(a^2 - b^2)`

⇒ `(x + 1)/(x - 1) = (sqrt(a^2 + b^2))/(sqrt(a^2 - b^2)`

Squaring both sides, we have

⇒ `(x + 1)^2/(x - 1)^2 = (a^2 + b^2)/(a^2 - b^2)`

⇒ `(x^2 + 2x + 1)/(x^2 - 2x + 1) = (a^2 + b^2)/(a^2 - b^2)`

Applying componendo and dividendo both sides, we get

⇒ `((x^2 + 2x + 1) + (x^2 - 2x + 1))/((x^2 + 2x + 1) - (x^2 - 2x + 1)) = (a^2 + b^2 + a^2 - b^2)/(a^2 + b^2 - a^2 - b^2)`

⇒ `(2x^2 + 2)/(4x) = (2a^2)/(2b^2)`

⇒ `(2(x^2 + 1))/(4x) = a^2/b^2`

⇒ `(x^2 + 1)/(2x) = a^2/b^2`

⇒ b²(x² + 1) = 2a²x

⇒ `b^2 = (2a^2x)/(x^2 + 1)`