A) \[2\tan \alpha \]
B) \[1\]
C) \[2\]
D) \[{{\sec }^{2}}\alpha \]
Correct Answer: A
Solution :
Here,\[\alpha \in \left[ 0,\,\,\frac{\pi }{2} \right)\] \[\Rightarrow \]\[\tan \alpha \]is\[(+ve)\] As, we know, if\[a,\,\,b>0\Rightarrow \frac{a+b}{2}\ge \sqrt{ab}\] i.e., \[AM\ge GM\] \[\therefore \]\[\frac{\sqrt{{{x}^{2}}+x}+\frac{{{\tan }^{2}}\alpha }{\sqrt{{{x}^{2}}+x}}}{2}\ge \sqrt{\sqrt{{{x}^{2}}+x}\cdot \frac{{{\tan }^{2}}\alpha }{\sqrt{{{x}^{2}}+x}}}\] (using\[AM\ge GM\]) \[\Rightarrow \] \[\sqrt{{{x}^{2}}+x}+\frac{{{\tan }^{2}}\alpha }{\sqrt{{{x}^{2}}+x}}\ge 2\tan \alpha \]
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