Question:
If $\sec x=t+\frac{1}{4 t}$, then the value of $\sec x+\tan x$ is ______________ .
Solution:
Given sec $x=t+\frac{1}{4 t}$
Since $\tan ^{2} x=\sec ^{2} x-1$
$=\left(t+\frac{1}{4 t}\right)^{2}-1$
$=t^{2}+\frac{1}{16 t^{2}}+\frac{1}{2}-1$
$=t^{2}+\frac{1}{16 t^{2}}-\frac{1}{2}$
$\tan ^{2} x=\left(t-\frac{1}{4 t}\right)^{2}$
i. e $\tan x=\pm\left(t-\frac{1}{4 t}\right)$
$\therefore \sec x+\tan x=\left(t+\frac{1}{4 t}\right) \pm\left(t-\frac{1}{4 t}\right)$
Hence, sec $x+\tan x=2 t$ or $\frac{1}{2 t}$
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