If -0- -2 then -x2-x-tan 2-x2-x is always greater than or equal toA- 2 tan-B- 2-C- 1D- 2

The correct option is A (a) √(x^2 + x) + tan^2 α/√(x^2 + x)≥ 2 tan α (A.M ≤ G.M). ...

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A.M Greater than or Equal to G.M

If α∈0, π/2 t...

Question

If $α \in (0, \frac{π}{2})$ then $\sqrt{x^{2} + x} + \frac{t a n^{2} α}{\sqrt{x^{2} + x}}$ is always greater than or equal to

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$I f α ϵ (0, \frac{π}{2}), t h e n \sqrt{x^{2} + x + \frac{t a n^{2} α}{\sqrt{x^{2} + x}}}$ is always greater than or equal to

α \in (0, \frac{π}{2})

\sqrt{x^{2} + x} + \frac{{tan}^{2} α}{\sqrt{x^{2} + x}}

(x \neq 0, - 1)

α \in (0, \frac{π}{2})

\sqrt{x^{2} + x} + \frac{{tan}^{2} α}{\sqrt{x^{2} + x}}

α ϵ [0, \frac{π}{2})

\sqrt{x^{2} + x} + \frac{{tan}^{2} α}{\sqrt{x^{2} + x}}

α \in (0, \frac{π}{2})

\sqrt{x^{2} + x} + \frac{{tan}^{2} α}{\sqrt{x^{2} + x}}

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