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Question
For the principal value, evaluate of the following:
`tan^-1{2sin(4cos^-1 sqrt3/2)}`
Solution
`tan^-1{2sin(4cos^-1 sqrt3/2)} = tan^-1{2sin[4cos^-1(cos pi/6)]}`
`=tan^-1{2sin[4xxpi/6]}`
`=tan^-1(2sin (2pi)/3)`
`=tan^-1[2xx(sqrt3/2)]`
`=tan^-1(sqrt3)`
`=tan^-1[tan(pi/3)]`
`= pi/3`
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