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Question
Find the value of `tan^-1 (tan (5pi)/6) +cos^-1(cos (13pi)/6)`
Solution
We know that `(5pi)/6 ∉ (- pi/2, pi/2)` and `(13pi)/6 ∉ [0, pi]`
∴ `tan^-1 (tan (5pi)/6) + cos^1(cos (13pi)/6)`
= `tan^-1 [tan (pi - pi/6)] + cos^-1[cos(2pi + pi/6)]`
= `tan^-1[tan(- pi/6)] + cos^-1(cos pi/6)`
= `tan^-1 (tan pi/6)+ cos^-1 (cos pi/6)`
= `- tan^-1 (tan pi/6) + cos^-1(cos pi/6)` .....[∵ tan–1(– x) = – tan– 1x]
= `- pi/6 + pi/6`
= 0
Hence, `tan^-1 (tan (5pi)/6) +cos^-1(cos (13pi)/6)` = 0
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