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प्रश्न
Evaluate the following:
`sin(sec^-1 17/8)`
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उत्तर
`=sin(sec^-1 17/8)=sin(cos^-1 8/17)`
`=sin[sin^-1sqrt(1-(8/17)^2)]` `[thereforecos^-1x=sin^-1sqrt(1-x^2)]`
`=sin[sin^-1(sqrt(1-64/289))]`
`=sin[sin^-1(sqrt(225/289))]`
`=sin[sin^-1 15/17]`
`=15/17`
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