Question

If $\sec \theta =m$ and $\tan \theta =n$ , then $\frac{1}{m}\left[(m+n)+\frac{1}{(m+n)}\right]$ is :

• A. 2
• B. 2 m
• C. 2 n
• D. mn

Answer 1

Answer: (A)

Given that $\sec \theta =m\tan \theta =n$
$\therefore \frac{1}{m}\left[\left(m+n\right)+\frac{1}{\left(m+n\right)}\right]$
$=\frac{1}{\sec \theta }\left[\sec \theta +\tan \theta +\frac{1}{\sec \theta +\tan \theta }\right]$
$=\frac{\left[{\sec }^2\theta +{\tan }^2\theta +2\sec \theta \tan \theta +1\right]}{\sec \theta \left(\sec \theta +\tan \theta \right)}$
$=\frac{2{\sec }^2\theta +2\sec \theta \tan \theta }{\sec \theta \left(\sec \theta +\tan \theta \right)}$
$=\frac{2\sec \theta \left(\sec \theta +\tan \theta \right)}{\sec \theta \left(\sec \theta +\tan \theta \right)}$
$=2$