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Question
Find the value of the following:
`tan^-1 [2 cos (2 sin^-1 1/2)]`
Solution
`tan^-1 [2 cos (2 sin^-1 1/2)]`
`Rightarrow sin^-1 1/2 = x`
`Rightarrow sin x = 1/2 = sin (pi/6)`
`therefore sin^-1 pi/6`
`Rightarrow tan^-1 [2 cos (2 xx sin^-1 1/2)]`
`= tan^-1 [2 cos (2 xx pi/6)]`
`Rightarrow tan^-1 [2 cos (pi/3)]`
`= tan^-1 [2 xx 1/2]`
`Rightarrow tan^-1 1`
`= pi/4`
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