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Question
Find the value of the following:
`tan^(-1)(1) + cos^(-1) (-1/2) + sin^(-1) (-1/2)`
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Solution
Let tan−1(1) = x
Then tan x = 1 = `tan pi/4`
∴ `tan^(-1) (1) = pi/4`
Let `cos^(-1) (-1/2)` = y
Then, cos y = `-1/2 = -cos(pi/3) = cos(pi - pi/3) = cos ((2pi)/3)`
∴ `cos^(-1) (- 1/2) = (2pi)/3`
Let `sin^(-1) (-1/2)` = z
Then sin z = `-1/2 = -sin(pi/6) = sin(-pi/6)`
∴ `sin^(-1)(-1/2) = - pi/6`
∴ `tan^(-1) (1) + cos^(-1) (-1/2) + sin^(-1) (-1/2)`
= `pi/4 + (2pi)/3 - pi/6`
= `(3pi + 8pi - 2pi)/12 `
= `(9pi)/12`
= `(3pi)/4`
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