1.

If `x in [-1, 0)` then find the value of `cos^(-1) (2x^(2) -1) -2 sin^(-1) x`

Answer» Correct Answer - `pi`
`cos^(-1) (2x^(2) -1) = 2pi - 2 cos^(-1) x " " ("as " x lt 0)`
`rArr cos^(-1) (2x^(2) -1) -2 sin^(-1) x = 2pi - 2 cos^(-1) x - 2sin^(-1) x`
`= 2pi - 2 (cos^(-1) x + sin^(-1) x)`
`= 2pi - 2 (pi)/(2) = pi`


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