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प्रश्न
`sin^-1{(sin - (17pi)/8)}`
उत्तर
We know
`sin(sin^-1theta)=theta if - pi/2<=theta<=pi/2`
We have
`sin^-1(sin - (17pi)/8)sin^-1(-sin (17pi)/8)`
`=sin^-1{-sin(2pi+pi/8)}`
`=sin^-1(-sin pi/8)`
`=sin^-1(sin-pi/8)`
`=-pi/8`
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