Question

\[ \text{If } \sec x=t+\frac1{4t}, \]

\[ \text{then the value of } \sec x+\tan x \text{ is} \]

Solution

Let

\[ \sec x+\tan x=y \]

Using identity

\[ (\sec x+\tan x)(\sec x-\tan x)=1 \]

\[ \sec x-\tan x=\frac1y \]

Adding,

\[ 2\sec x = y+\frac1y \]

Given,

\[ \sec x=t+\frac1{4t} \]

\[ 2\left(t+\frac1{4t}\right) = y+\frac1y \]

\[ 2t+\frac1{2t} = y+\frac1y \]

Comparing,

\[ y=2t \]

Therefore,

\[ \sec x+\tan x=2t \]

Answer

\[ \boxed{2t} \]