Q94 / 180 · COMEDK 2016 · Mathematics
If $\sec \theta=m$ and $\tan \theta=n$, then $\frac{1}{m}\left[(m+n)+\frac{1}{(m+n)}\right]=$
- A$m n$
- B$2 n$
- C$2 m$
- Correct answerD2
Solution
We have, $\sec \theta=m$ and $\tan \theta=n$
Now, $\frac{1}{m}\left[(m+n)+\frac{1}{(m+n)}\right]$
$=\frac{1}{\sec \theta}\left[(\sec \theta+\tan \theta)+\frac{1}{(\sec \theta+\tan \theta)}\right]$
$=\frac{1}{\sec \theta}\left[\frac{(\sec \theta+\tan \theta)^{2}+1}{(\sec \theta+\tan \theta)}\right]$
$=\frac{1}{\sec \theta}\left[\frac{\sec ^{2} \theta+\tan ^{2} \theta+2 \sec \theta \tan \theta+1}{(\sec \theta+\tan \theta)}\right]$
$=\frac{1}{\sec \theta}\left[\frac{\sec ^{2} \theta+\sec ^{2} \theta-1+2 \sec \theta+\tan \theta+1}{\sec \theta+\tan \theta}\right]$
$=\frac{1}{\sec \theta}\left[\frac{2 \sec 2 \theta+2 \sec \theta \tan \theta}{\sec \theta+\tan \theta}\right]$
$=\frac{1}{\sec \theta}\left[\frac{2 \sec \theta(\sec \theta+\tan \theta)}{(\sec \theta+\tan \theta)}\right]$
$=\frac{1}{\sec \theta}(2 \sec \theta)=2$
Hence, the correct answer is option: 'D'.