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