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Question
Find the value of the following:
`tan^(-1) (tan (7x)/6)`
Solution 1
We know that tan−1 (tan x) = x if `x in (-pi/2,pi/2)`, which is the principal value branch of tan −1x.
Here `(7pi)/6 !in (-pi/2, pi/2)`
Now `tan^(-1) (tan (7pi)/6)` can be written as
`tan^(-1) (tan (7pi)/6) = tan^(-1) [tan(2pi - (5pi)/6)]` `[tan(2pi - x) = - tan x]`
`= tan^(-1) [-tan ((5pi)/6)] `
`= tan^(-1) [tan ((-5pi)/6)]`
` = tan^(-1) [tan(pi - (5pi)/6)]`
`= tan^(-1) [tan(pi/6)], " where" pi/6 in (-pi/2, pi/2)`
`:. tan^(-1) (tan (7pi)/6)`
` = tan^(-1) (tan pi/6) = pi/6`
Solution 2
Given, `tan^-1(tan (7pi)/6)`
We know that, for x ∈ `(-pi/2, pi/2)`, `cos^-1(cosx) = x`
= `tan^-1(tan (7pi)/6)`
`= tan^-1(tan(pi + pi/6))`
= `tan^-1(tan pi/6)`
`= pi/6`
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