JEE Advance - Mathematics (2003 - No. 20)
If $$\,\alpha \in \left( {0,{\pi \over 2}} \right)\,\,then\,\,\sqrt {{x^2} + x} + {{{{\tan }^2}\alpha } \over {\sqrt {{x^2} + x} }}$$ is always greater than or equal to
$$2\,\tan \alpha \,$$
1
2
$${\sec ^2}\,\alpha $$