Question

If `cos^-1  x/2+cos^-1  y/3=alpha,` then prove that  `9x^2-12xy cosa+4y^2=36sin^2a.`

Answer 1

We know

`cos^-1x+cos^-1y=cos^-1[xy-sqrt(1-x^2)sqrt(1-y^2)]`

Now,

`cos^-1  x/2+cos^-1  y/3=alpha,`

⇒ `cos^-1[x/2  y/3-sqrt(1-x^2/4)sqrt(1-y^2/3)]=alpha`

⇒ `x/2  y/3-sqrt(1-x^2/4)sqrt(1-y^2/3)=cos alpha`

⇒ `xy-sqrt(4-x^2)sqrt(9-y^2)=6cosalpha`

⇒ `sqrt(4-x^2)sqrt(9-y^2)=xy-6cosalpha`

⇒ `(4-x^2)(9-y^2)=x^2y^2+36cos^2alpha-12xycosalpha`      [Squaring both sides]

⇒ `36-4y^2-9x^2+x^2y^2=x^2y^2+36cos^2alpha-12xycosalpha`

⇒ `36-4y^2-9x^2+36cos^2alpha-12xycosalpha`

⇒ `9x^2-12xy  cosalpha+4y^2=36-36cos^2alpha`

⇒ `9x^2-12xy  cosalpha+4y^2=36sin^2alpha`