Question

If a $a \in \bigg(0,\frac{\pi}{2}\bigg),$ then $\sqrt {x^2+x} +\frac{\tan^2\, a}{\sqrt {x^2+x}}$ is always greater than or equal to

• A. $2\, \tan\, a$
• B. 1
• C. 2
• D. $\sec^2 a$

Answer 1

Answer: (A)

$2\, \tan\, a$

Answer 2

Here, $a \in \, (0,\frac{\pi}{2}) \Rightarrow \tan\, a > 0$
$\therefore \, \, \, \, \, \, \frac{\sqrt {x^2+x} +\frac{\tan^2\, a}{\sqrt {x^2+x}} }{2} \ge \sqrt{\sqrt {x^2+x} .\frac{\tan^2\, a}{\sqrt {x^2+x}} }$
\hspace60mm [ using $AM \ge GM]$
$\Rightarrow \, \, \, \sqrt {x^2+x} +\frac{\tan^2\, a}{\sqrt {x^2+x}} \ge 2 \, \tan \, a$

The correct option is(A): 2 tan a