Solve that following equation: `cottheta+t a ntheta=2`

Explanation

`cottheta+tantheta=2`
`costheta/sintheta+sintheta/costheta=2`
Taking LCM
`(cos^2theta+sin^2theta)/(sinthetacostheta)=2`
Using identity
`cos^2theta+sin^2theta=1`
`1/(costhetasintheta)=2`
`implies 2costhetasintheta=2`
`implies sin2theta=1`
`sin2theta=sin(pi/2)`
Hence, the general solution is given by,
`2theta=npi+(-1)^npi/2`, where `n in Z`
`theta=(npi)/2+(-1)^npi/4`.