Question

If $\sec x=t+\frac{1}{4t},$ then the value of sec x + tan x is _________.

Answer 1

$$\begin{gathered} \mathrm{Given}\sec x=t+\frac{1}{4t} \\ \mathrm{Since}{\tan }^{2}x={\sec }^{2}x-1 \\ ={\left(t+\frac{1}{4t}\right)}^2-1 \\ =t^2+\frac{1}{16t^2}+\frac{1}{2}-1 \\ =t^2+\frac{1}{16t^2}-\frac{1}{2} \\ {\tan }^{2}x={\left(t-\frac{1}{4t}\right)}^2 \\ \mathrm{i}.\mathrm{e}\tan x=\pm \left(t-\frac{1}{4t}\right) \\ \therefore \sec x+\tan x=\left(t+\frac{1}{4t}\right)\pm \left(t-\frac{1}{4t}\right) \\ \mathrm{Hence},\sec x+\tan x=2t\mathrm{or}\frac{1}{2t} \end{gathered}$$