Question

If $\sec$ $\beta =\alpha +\frac{1}{4a}$,then the value of $\sec \beta +\tan \beta$ is

• A. $a$ or $\frac{1}{a}$
• B. $2a$ or$\frac{1}{2a}$
• C. $4a$ or $\frac{1}{4a}$
• D. $1$

Answer 1

Answer: (B)

$2a$ or$\frac{1}{2a}$

We know that

${\sec }^2\beta -{\tan }^2\beta =1$

Therefore

$\tan \beta =\pm \sqrt{{\sec }^2\beta -1}$

$\tan \beta =\pm \sqrt{{\alpha }^2+\left(\frac{1}{4\alpha }\right){}^2+\frac{1}{2}}-1$

$\tan \beta =\pm \sqrt{{\alpha }^2+\left(\frac{1}{4\alpha }\right){}^2-\frac{1}{2}}$

$\tan \beta =\pm \sqrt{(\alpha -\frac{1}{4\alpha }){}^2}$

$\tan \beta =\pm (\alpha -\frac{1}{4\alpha })$ ...(i)

Hence $sec\alpha +\tan \alpha =2\alpha or\frac{1}{2\alpha }$

The answer is B