Question

If sec x = m and tan x = n, then $\frac{1}{m}\left\{\left(m+n\right)+\frac{1}{m+n}\right\}$ is equal to ____________.

Answer 1

If sec x = m and tan x = n
$$\begin{gathered} \frac{1}{m}\left\{\left(m+n\right)+\frac{1}{m+n}\right\} \\ \mathrm{i}.\mathrm{e}\frac{1}{\sec x}\left\{\sec x+\tan x+\frac{1}{\sec x+\tan x}\right\} \\ =\frac{1}{\sec x}\left\{\sec x+\tan x+\frac{1}{\sec x+\tan x}\times \left(\frac{\sec x-\tan x}{\sec x-\tan x}\right)\right\} \\ =\frac{1}{\sec x}\left\{\sec x+\tan x+\frac{\sec x-\tan x}{{\sec }^{2}x-{\tan }^{2}x}\right\} \\ =\frac{1}{\sec x}\left\{\sec x+\tan x+\frac{\sec x-\tan x}{1}\right\} \\ =\frac{1}{\sec x}\left\{2\sec x\right\} \\ =2 \\ \therefore \frac{1}{m}\left\{\left(m+n\right)+\frac{1}{m+n}\right\}=2 \end{gathered}$$