Question

If `cos^-1  x/2 + cos^-1  y/3 = θ`, then prove that 9x² – 12xy cos θ + 4y² = 36 sin² θ

Answer 1

Given: `cos^-1  x/2 + cos^-1  y/3 = θ`

⇒ `cos^-1 [x/2 xx y/3 - sqrt(1 - x^2/4) * sqrt(1 - y^2/9)] = θ`  ...`[∵ cos^-1a + cos^-1b = cos^-1 (ab - sqrt(1 - a^2) * sqrt(1 - b^2))]`

⇒ `(xy)/6 - sqrt(1 - x^2/4) * sqrt(1 - y^2/9) = cos θ`

⇒ `(xy)/6 - cos θ = sqrt(1 - x^2/4) * sqrt(1 - y^2/9)`

Squaring both sides, we get

⇒ `((xy)/6 - cos θ)^2 = (1 - x^2/4)(1 - y^2/9)`

⇒ `(x^2y^2)/36 + cos^2θ - (2xy)/6 cos θ = 1 - x^2/4 - y^2/9 + (x^2y^2)/36`

⇒ `x^2/4 + y^2/9 - (xy cos θ)/3 = 1 - cos^2θ`

⇒ `(9x^2 + 4y^2 - 12xy cos θ)/36 = sin^2 θ`

⇒ 9x² + 4y² – 12xy cos θ = 36 sin² θ

⇒ 9x² – 12xy cos θ + 4y² = 36 sin² θ

Hence Proved.