A) 4x
B) 2x
C) x
D) \[\frac{x}{2}\]
Correct Answer: B
Solution :
| Given, \[\sec \theta =x+\frac{1}{4x}\] |
| Squaring both sides, we get |
| \[{{\sec }^{2}}\theta ={{\left( x+\frac{1}{4x} \right)}^{2}}\] |
| \[={{x}^{2}}+\frac{1}{16{{x}^{2}}}+\frac{1}{2}\] |
| \[{{\sec }^{2}}\theta -1={{x}^{2}}+\frac{1}{16{{x}^{2}}}-\frac{1}{2}\] |
| \[{{\sec }^{2}}\theta -1={{\left( x-\frac{1}{4x} \right)}^{2}}\] |
| \[{{\tan }^{2}}\theta ={{\left( x-\frac{1}{4x} \right)}^{2}}\] |
| \[\left[ \because \,{{\sec }^{2}}\theta -1\,={{\tan }^{2}}\theta \right]\] |
| \[\tan \,\theta =\underline{+}\left( x-\frac{1}{4x} \right)\] |
| \[\sec \theta +\tan \theta \] |
| \[x+\frac{1}{4x}+x-\frac{1}{4x}=2x\] |
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