if `tantheta+sectheta=l ` then prove that `tantheta=(l^2+1)/(2l)`

Explanation

Given that
`tantheta+sectheta=l`
`rArrsectheta+tantheta=l`
We know that , " " `sec^(2)theta-tan^(2)theta=1`
`rArr(sectheta-tantheta)(sectheta+tantheta)=1`
`rArrsectheta=tantheta=(1)/(sectheta+tantheta)=(1)/(l)`
Adding equations (1) and (2).,we get
`2sectheta=l+(1)/(l)=(l^(2)+1)/(l)`
`rArrsectheta=(l^(2)+1)/(2l)`