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प्रश्न
`sin^-1{cos(sin^-1 sqrt3/2)}`
उत्तर
`sin^-1{cos(sin^-1 sqrt3/2)}=sin^-1{cos(sin^-1 sin pi/3)}`
`=sin^-1{cos(pi/3)}`
`=sin^-1{1/2}`
`=sin^-1{sin pi/6}`
`=pi/6`
APPEARS IN
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