If $\alpha \in \left(0,\frac{\pi }{2}\right),$ then $\sqrt{x^2+x}+\frac{{\tan }^2\alpha }{\sqrt{x^2+x}}$ is always greater than or equal to $\left(x\ne 0,-1\right)$

• A

  2

• B

  1

• C

  $2\tan \alpha$

• D

  $2{\sec }^2\alpha$