Question

If cot θ + cos θ = p and cot θ − cos θ = q, prove that `p^2 - q^2 = 4sqrt(pq)`.

Answer 1

p = cot θ + cos θ

= `cos θ/sin θ + cos θ`

= `(cos θ(1 + sin θ))/sin θ`

q = cot θ − cos θ

= `(cos θ(1 - sin θ))/sin θ`

⇒ p² − q² = `((cos θ(1 + sin θ))/sin θ)^2 - ((cos θ(1 - sin θ))/sin θ)^2`

= `(cos^2 θ(1 + sin θ)^2)/(sin^2 θ) - (cos^2 θ(1 - sin θ)^2)/sin^2 θ`

= `(cos^2 θ)/(sin^2 θ)` [1 + sin² θ + 2sin θ − 1 − sin² θ + 2 sin θ]

= `(cos^2 θ)/(sin^2 θ) xx 4sin θ`

= `(4 cos^2 θ)/sin θ`

⇒ p × p = `(cos θ(1 + sin θ))/(sin θ) xx (cos θ(1 - sin θ))/sin θ`

= `(cos^2 θ)/(sin^2 θ) (1 - sin^2 θ)`

= `(cos^2 θ)/(sin^2 θ) cos^2 θ`

⇒ pq = `(cos^4 θ)/(sin^ 2 θ)`

`sqrt(pq) = cos^2θ/sinθ`

`4sqrt(pq) = (4cos^2θ)/sinθ - p^2 - q^2`

∴ `p^2 - q^2 = 4sqrt(pq)`

∴ Hence proved.