Question

If cos^-1 x/2 + cos^-1 y/3 = alpha then prove that 9x^2 -12xycos alpha + 4y^2 = 36 sin^2 alpha

Answer 1

Given, cos⁻¹(x/2) + cos⁻¹(y/3) = α
take cos both sides,
cos[cos⁻¹(x/2) + cos⁻¹(y/3)] = cosα
⇒cos(cos⁻¹(x/2)) .cos(cos⁻¹(y/3)) - sin(cos⁻¹(x/2)).sin(cos⁻¹(y/3)) = cosα
⇒x/2.y/3 - √(1 - x²/4).√(1 - y²/9) = cosα
⇒xy/6 - cosα = √(1 - x²/4 - y²/9 + x²y²/36)
Squaring both sides,
(xy/6 - cosα)² = 1 - x²/4 - y²/9 + x²y²/36
⇒ x²y²/36 + cos²α - xycosα/3 = 1 - x²/4 - y ²/9 + x²y²/36
⇒( 3cos²α - xycosα)/3 = (36 - 9x² - 4y²)/36
⇒36cos²α - 12xycosα = 36 - 9x² - 4y²
⇒ 36 - 36sin²α - 12xycosα = 36 - 9x² - 4y²
⇒9x² + 4y² - 12xycosθ = 36sin²α
Hence proved

Answer 2

Answer:

Step-by-step explanation: