If sec θ + tan θ = x, then sec θ =

Given: $\sec \theta+\tan \theta=x$

We know that,

$\sec ^{2} \theta-\tan ^{2} \theta=1$

$\Rightarrow(\sec \theta+\tan \theta)(\sec \theta-\tan \theta)=1$

$\Rightarrow x(\sec \theta-\tan \theta)=1$

$\Rightarrow \sec \theta-\tan \theta=\frac{1}{x}$

Now,

$\sec \theta+\tan \theta=x$

$\sec \theta-\tan \theta=\frac{1}{x}$

Adding the two equations, we get

$(\sec \theta+\tan \theta)+(\sec \theta-\tan \theta)=x+\frac{1}{x}$

$\Rightarrow \sec \theta+\tan \theta+\sec \theta-\tan \theta=\frac{x^{2}+1}{x}$

$\Rightarrow 2 \sec \theta=\frac{x^{2}+1}{x}$

$\Rightarrow \sec \theta=\frac{x^{2}+1}{2 x}$

Therefore, the correct choice is (b).