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प्रश्न
`sin^-1x=pi/6+cos^-1x`
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उत्तर
`sin^-1x=pi/6+cos^-1x`
⇒ `sin^-1x=pi/6+pi/2-sin^-1x` `[becausecos^-1x=pi/2-sin^-1x]`
⇒ `2sin^-1x=(2pi)/3`
⇒ `sin^-1x=pi/3`
⇒ `sin^-1x=pi/3`
⇒ `x=sin pi/3=sqrt3/2`
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