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If cos^−1(xa)+cos^−1(yb)=α , prove that x2/a2−2 xy/ab cosα+y^2/b^2=sin^2 α - Mathematics

If `cos^-1( x/a) +cos^-1 (y/b)=alpha` , prove that `x^2/a^2-2(xy)/(ab) cos alpha +y^2/b^2=sin^2alpha`

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Solution

`cos^-1( x/a) +cos^-1 (y/b)=alpha` 

`cos^-1(x/a) =alpha-cos^-1 (y/b)`

`=>cos {cos^-1 (x/a)}=cos{alpha-cos^-1 (y/b)}`

`=>x/a=cos alpha cos{cos^-1 (y/b)}+sinalpha sin{cos^-1(y/b)}`

`=>x/a=y/b cos alpha+sin alpha sqrt(1-(y/b)^2)`

`=>x/a - y/b cos alpha =sin alpha sqrt(1-(y/b)^2)`

Squaring both sides, we get

`(x/a - y/b cos alpha)^2={sin alpha sqrt(1-(y/b)^2)}^2`

`(x/a)^2+(y/b)^2cos^2 alpha- (2xy)/(ab) cos alpha=sin^2 alpha- sin^2 alpha(y/b)^2`

`(x/a)^2+(y/b)^2cos^2 alpha+sin^2 alpha(y/b)^2-(2xy)/(ab) cos alpha=sin^2 alpha`

`(x/a)^2+(y/b)^2(cos^2 alpha+sin^2 alpha)-(2xy)/(ab) cos alpha=sin^2 alpha`

`=>(x/a)^2-(2xy)/(ab) cos alpha+(y/b)^2=sin^2 alpha`

Concept: Inverse Trigonometric Functions (Simplification and Examples)
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