Question

If $\sec \theta =a+\frac{1}{4a}then\sec \theta +\tan \theta =$

• A. a
• B. 2a
• C. 3a
• D. 4a

Answer 1

The correct option is B 2a

$\sec \theta +\tan \theta =\sec \theta +\sqrt{{\sec }^2\theta -1}=a+\frac{1}{4a}+\sqrt{(a+\frac{1}{4a}{)}^2-1}=a+\frac{1}{4a}+a-\frac{1}{4a}+a-\frac{1}{4a}=2a$