If $\sec\theta +\tan\theta =11/3\,then\,\sec\theta =$

Given that:-

$\sec \theta+\tan \theta=\frac{11}{3} \dots(i)$

we know that

$\begin{aligned} & \sec ^{2} \theta-\tan ^{2} \theta=1 \\ & (\sec \theta+\tan \theta)(\sec \theta-\tan \theta)=1 \\ \Rightarrow & \frac{11}{3} \cdot(\sec \theta-\tan \theta)=1 \\ \Rightarrow & \sec \theta-\tan \theta=\frac{3}{11} \dots(ii) \end{aligned}$

Adding eqn (i) and (ii), we get

$\begin{array}{l} \sec \theta+\tan \theta=\frac{11}{3} \\ \sec \theta-\tan \theta=\frac{3}{11} \\ 2 \sec \theta=\frac{11}{3}+\frac{3}{11} \\ \Rightarrow 2 \sec \theta=\frac{121+9}{33} \\ \Rightarrow 2 \sec \theta=\frac{130}{33} \\ \Rightarrow \sec \theta=\frac{130}{2 \times 33}=\frac{65}{33} \\ \therefore \sec \theta=\frac{65}{33} \end{array}$