Find the value of:
i) `sin ^(-1)(sin (3pi)/5)`, ii) `cos^(-1)(cos(13pi)/6)`, iii) `tan^(-1)(tan (7pi)/6)`.

Explanation

i) We know that the principal-value branch of `cos^(-1)` is `[0,pi]`.
`therefore cos^(-1)(cos (13pi)/6) ne (13pi)/6`.
Now, `cos^(-1)(cos (13pi)/6=cos^(-1){cos(2pi+pi/6)}=cos^(-1){cospi/6}[therefore cos(2pi+theta)=costheta]=pi/6`.
Hence, `cos^(-1)(cos (13pi)/6)) =pi/6`.
iii) We know that the principal-value branch of `tan^(-1)` is `(-pi/, pi/2)`.
`therefore tan^(-1)(tan(7pi)/6)) ne (7pi)/6`.
Now, `tan^(-1)(tan(7pi)6) = tan^(-1){tan(pi+pi/6)}`
`=tan^(-1)(tanpi/6) [therefore tan(pi+theta)=tantheta]`
`=pi/6`.
Hence, `tan^(-1)(tan(7pi)/6)=pi/6`.