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)

2

Answer 2

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