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If `Cos^-1 X/2+Cos^-1 Y/3=Alpha,` Then Prove That `9x^2-12xy Cosa+4y^2=36sin^2a.` - Mathematics

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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.`

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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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पाठ 4: Inverse Trigonometric Functions - Exercise 4.13 [पृष्ठ ९२]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 4 Inverse Trigonometric Functions
Exercise 4.13 | Q 1 | पृष्ठ ९२

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