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प्रश्न
Prove `(9pi)/8 - 9/4 sin^(-1) 1/3 = 9/4 sin^(-1) (2sqrt2)/3`
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उत्तर
L.H.S = `(9pi)/8 - 9/4 sin^(-1) 1/3`
`= 9/4 (pi/2 - sin^(-1) 1/3)`
`= 9/4 (cos^(-1) 1/3)` ....(1) `[sin^(-1)x + cos^(-1) x = pi/2]`
Now, let `cos^(-1) 1/3 = x` Then, `cos x = 1/3 => sin x = sqrt(1 - (1/3)^2) = (2sqrt2)/3`
`:. x = sin^(-1) (2sqrt2)/3 => cos^(-1) 1/3 = sin^(-1) (2sqrt2)/3`
:. L.H.S = `9/4 sin^(-1) (2(sqrt2))/3` = R.H.S
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