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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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