A) \[2\]
B) \[2m\]
C) \[2n\]
D) \[mn\]
Correct Answer: A
Solution :
Given that,\[sec\theta =m\]and\[\tan \theta =n\] \[\therefore \] \[\frac{1}{m}\left[ (m+n)+\frac{1}{(m+n)} \right]\] \[=\frac{1}{\sec \theta }\left[ \sec \theta +\tan \theta +\frac{1}{\sec \theta +\tan \theta } \right]\] \[=\frac{[{{\sec }^{2}}\theta +{{\tan }^{2}}\theta +2\sec \theta \tan \theta +1]}{\sec \theta (\sec \theta +\tan \theta )}\] \[=\frac{2\sec \theta +2\sec \theta \tan \theta }{\sec \theta (\sec \theta +\tan \theta )}\] \[=\frac{2\sec \theta (\sec \theta +\tan \theta )}{\sec \theta (\sec \theta +\tan \theta )}\] \[=2\]
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