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प्रश्न
`(9pi)/8-9/4sin^-1 1/3=9/4sin^-1 (2sqrt2)/3`
उत्तर
`(9pi)/8-9/4sin^-1 1/3=9/4sin^-1 (2sqrt2)/3`
LHS = `(9pi)/8-9/4sin^-1 1/3`
`=9/4(pi/2-sin^-1 1/3)`
`=9/4(cos^-1 1/3)`
`=9/4(sin^-1sqrt(1-1/9))`
`=9/4(sin^-1 (2sqrt2)/3)=` RHS
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