Given. $\tan \theta+\sec \theta=l$ …..$(i)$

[multiply by $(\sec \theta-\tan \theta)$ on numerator and denominator L.H.S.]

$\Rightarrow \quad \frac{(\tan \theta+\sec \theta)(\sec \theta-\tan \theta)}{(\sec \theta-\tan \theta)}=l \Rightarrow \frac{\left(\sec ^{2} \theta-\tan ^{2} \theta\right)}{(\sec \theta-\tan \theta)}=l$

$\Rightarrow$ $\frac{1}{\sec \theta-\tan \theta}=t$ $\left[\because \sec ^{2} \theta-\tan ^{2} \theta=1\right]$

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

On adding Eqs. $(i)$ and $(ii),$ we get

$2 \sec \theta=l+\frac{1}{l}$

$\sec \theta=\frac{l^{2}+1}{2 l}$ Hence proved.