Question

\[ \text{If } \sec x=m \text{ and } \tan x=n, \]

\[ \frac1m\left\{(m+n)+\frac1{m+n}\right\} \]

\[ \text{is equal to} \]

Solution

Given,

\[ m=\sec x,\quad n=\tan x \]

So,

\[ m+n=\sec x+\tan x \]

Using identity

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

\[ \frac1{m+n} = \sec x-\tan x \]

Therefore,

\[ (m+n)+\frac1{m+n} \]

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

\[ =2\sec x \]

Hence,

\[ \frac1m\left\{(m+n)+\frac1{m+n}\right\} = \frac1{\sec x}(2\sec x) =2 \]

Answer

\[ \boxed{2} \]