Consider the given equation.
secθ+tanθ=x ……. (1)
(secθ+tanθ)×secθ−tanθsecθ−tanθ=x
sec2θ−tan2θsecθ−tanθ=x
We know that
sec2θ−tan2θ=1
Therefore,
1secθ−tanθ=x
secθ−tanθ=1x …….. (2)
From equation (1) and (2), we get
2secθ=x+1x
2secθ=1+x2x
secθ=1+x22x
Hence, the value of secθ is 1+x22x.