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