If $\tan \theta + \sec \theta = l$,then prove that $\sec \theta = \frac{l^{2} + 1}{2l}$.

(A) Given: $\tan \theta + \sec \theta = l$ .....$(i)$
We know the identity: $\sec^{2} \theta - \tan^{2} \theta = 1$.
This can be written as $(\sec \theta - \tan \theta)(\sec \theta + \tan \theta) = 1$.
Substituting the value of $(\sec \theta + \tan \theta)$ from equation $(i)$:
$(\sec \theta - \tan \theta) \cdot l = 1$
$\sec \theta - \tan \theta = \frac{1}{l}$ .....$(ii)$
Now,adding equation $(i)$ and equation $(ii)$:
$(\tan \theta + \sec \theta) + (\sec \theta - \tan \theta) = l + \frac{1}{l}$
$2 \sec \theta = \frac{l^{2} + 1}{l}$
Dividing both sides by $2$:
$\sec \theta = \frac{l^{2} + 1}{2l}$.
Hence,it is proved.