Evaluate: `tan^(-1)(tan(5pi)/6)+cos^(-1){cos((13pi)/6)}`

Answer

N/a

Explanation

We know that, `tan^(-1)tanx = x, x in (-(pi)/(2), (pi)/(2))` and `cos^(-) x = x, x in [0,pi]`
`:. tan^(-1)(tan'(5pi)/(6)) + cos^(-1) (cos'(13pi)/(6))`
` = tan^(-1)[tan'(pi - (pi)/(6))] + cos^(-1)[cos'(pi+(7pi)/(6))]`
`= tan^(-1)(-tan'(pi)/(6)) + cos^(-1) (-cos'(7pi)/(6)) , [:' cos(pi+theta) = - cos theta]`
`= -tan^(-1) (tan'(pi)/(6)) + pi- [cos^(-1)cos'((7pi)/(6))]`
`{ :'tan^(-1)(-x)=-tan^(-1)x,x in R` and `cos^(-1)(-x) = pi - cos^(-1)x, x in [-1,1]`}
` = -tan^(-1)(tan'(pi)/(6)) + pi - cos^(-1)[cos(pi + (pi)/(6))]`
`= -tan^(-1)(tan'(pi)/(6)) + pi - [cos^(-1)(-cos'(pi)/(6))] , [:' cos (pi + theta) = - cos theta]`
`= - tan^(-1)(tan'(pi)/(6)) + pi - pi + cos^(-1)(cos'(pi)/(6)), [:' cos^(-1)(-x)=pi-cos^(-1)x]`
`= -(pi)/(6) + 0 + (pi)/(6) = 0`