A) \[\frac{{{k}^{2}}+1}{2k}\]
B) \[\frac{2k}{{{k}^{2}}+1}\]
C) \[\frac{k}{{{k}^{2}}+1}\]
D) \[\frac{k}{{{k}^{2}}-1}\]
E) none of these
Correct Answer: B
Solution :
\[\because \]\[\sec \theta +\tan \theta =k\] ...(i) \[\therefore \] \[\frac{{{\sec }^{2}}\theta -{{\tan }^{2}}\theta }{\sec \theta -\tan \theta }=k\] \[\Rightarrow \] \[\sec \theta -\tan \theta =\frac{1}{k}\] ...(ii) On solving Eqs (i) and (ii), we get \[2\sec \theta =k+\frac{1}{k}\] \[\Rightarrow \] \[\cos \theta =\frac{2k}{{{k}^{2}}+1}\]
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