A) \[{{m}^{2}}\]
B) \[\frac{1}{{{m}^{2}}}\]
C) \[\frac{1}{m}\]
D) None of these
Correct Answer: B
Solution :
| [b] \[\sec \theta +\tan \theta =m\] |
| \[\therefore \,\,\,\,\,\sec \theta -\tan \theta =\frac{1}{m}\] |
| Now, \[{{\sec }^{4}}\theta -{{\tan }^{4}}\theta -2\sec \theta \cdot \tan \theta \] |
| \[=({{\sec }^{2}}\theta -{{\tan }^{2}}\theta )({{\sec }^{2}}\theta +{{\tan }^{2}}\theta )-2\sec \theta \cdot \tan \theta \]\[=1\cdot ({{\sec }^{2}}\theta +{{\tan }^{2}}\theta )-2\sec \theta \cdot \tan \theta \] |
| \[={{(\sec \theta -\tan \theta )}^{2}}\]\[[{{a}^{2}}+{{b}^{2}}-2ab={{(a-b)}^{2}}]\] |
| \[=\frac{1}{{{m}^{2}}}\] |
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