Solve the following equation, `tan^(2)theta+sec^(2)theta+3 = 2(sqrt(2)sectheta+tantheta)`

Explanation

We have `tan^(2)theta+sec^(2)theta+3=2sqrt(2)sectheta+2tantheta`
`rArr tan^(2)theta+1-2tantheta+sec^(2)theta-2sqrt(2)sectheta+2=0`
`rArr (tantheta-1)^(2)+(sectheta-sqrt(2))^(2)=0 rArr tantheta=1` and `sectheta=sqrt(2)`
As the periodicity of `tantheta` and `sectheta` are not same, we get
`therefore =2npi+pi/4, n in I`