A) 2
B) 2m
C) 2n
D) mn
Correct Answer: A
Solution :
Given that \[\sec \theta =m\] and \[\tan \,\theta =n\] \[\therefore \] \[\frac{1}{m}\left[ (m+n)+\frac{1}{(m+n)} \right]\] \[=\frac{1}{\sec \theta }\left[ \sec \theta +\tan \theta +\frac{1}{\sec \theta +\tan \theta } \right]\] \[=\frac{[{{\sec }^{2}}\theta +{{\tan }^{2}}\theta +2\sec \theta \,tan\theta +1]}{\sec \theta (\sec \theta +\tan \theta )}\] \[=\frac{2{{\sec }^{2}}\theta +2\sec \theta \,\tan \theta }{\sec \theta (\sec \theta +\tan \theta )}\] \[=\frac{2\sec \theta (sec\theta +tran\theta )}{\sec \theta (\sec \theta +\tan \theta )}\] \[=2\]
You need to login to perform this action.
You will be redirected in
3 sec
