If $\sec x=m$ and $\tan x=n$, then $\frac{1}{m}\left\{(m+n)+\frac{1}{m+n}\right\}$ is equal to ____________.
If sec x = m and tan x = n
$\frac{1}{m}\left\{(m+n)+\frac{1}{m+n}\right\}$
i. e $\frac{1}{\sec x}\left\{\sec x+\tan x+\frac{1}{\sec x+\tan x}\right\}$
$=\frac{1}{\sec x}\left\{\sec x+\tan x+\frac{1}{\sec x+\tan x} \times\left(\frac{\sec x-\tan x}{\sec x-\tan x}\right)\right\}$
$=\frac{1}{\sec x}\left\{\sec x+\tan x+\frac{\sec x-\tan x}{\sec ^{2} x-\tan ^{2} x}\right\}$
$=\frac{1}{\sec x}\left\{\sec x+\tan x+\frac{\sec x-\tan x}{1}\right\}$
$=\frac{1}{\sec x}\{2 \sec x\}$
= 2
$\therefore \frac{1}{m}\left\{(m+n)+\frac{1}{m+n}\right\}=2$
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