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प्रश्न
If −1 < x < 0, then write the value of `sin^-1((2x)/(1+x^2))+cos^-1((1-x^2)/(1+x^2))`
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उत्तर
Let `x=-tany`
Where `0<y< pi/2`
Then,
`sin^-1((2x)/(1+x^2))+cos^-1((1-x^2)/(1+x^2))=sin^-1((-2tany)/(1+tan^2y))+cos^-1((1-tan^2y)/(1+tan^2y))`
`=sin^-1{-sin(2y)}+cos^-1{cos(2y)}`
`=-sin^-1{sin(2y)}+cos^-1{cos(2y)}`
`=-2y+2y`
= 0
`therefore sin^-1((2x)/(1+x^2))+cos^-1((1-x^2)/(1+x^2))=0`
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