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प्रश्न
For the principal value, evaluate the following:
`sin^-1[cos{2\text(cosec)^-1(-2)}]`
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उत्तर
`sin^-1[cos{2\text(cosec)^-1(-2)}]=sin^-1[cos{2\text(cosec)^-1\text(cosec- )pi/6}]`
`=sin^-1[cos{-pi/3}]`
`=sin^-1[cos(pi/3)]`
`=sin^-1(1/2)`
`=sin^-1(sin π/6)`
`=pi/6`
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