If \[\sec \theta +\tan \theta =2+\sqrt{3},\]then the value of\[\sec \theta \]is |
A) \[\sqrt{3}\]
B) \[2\]
C) \[4\]
D) \[2\sqrt{3}\]
Correct Answer: B
Solution :
\[\sec \theta +\tan \theta =2+\sqrt{3}\] |
On squaring both sides, we get |
\[{{\sec }^{2}}\theta +{{\tan }^{2}}\theta +2\sec \theta \tan \theta =4+3+4\sqrt{3}\] |
\[\Rightarrow \]\[{{\sec }^{2}}\theta +{{\sec }^{2}}\theta -1+2\sec \theta \tan \theta =7+4\sqrt{3}\] |
\[\Rightarrow \] \[2{{\sec }^{2}}\theta +2\sec \theta \tan \theta =8+4\sqrt{3}\] |
\[\Rightarrow \] \[2\sec \theta \,\,(\sec \theta +\tan \theta )=4\,\,(2+\sqrt{3)}\] |
\[\therefore \]\[\sec \theta =\frac{2\,\,(2+\sqrt{3})}{\sec \theta +\tan \theta }=\frac{2\,\,(2+\sqrt{3})}{(2+\sqrt{3})}=2\] |
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