If $\tan(\alpha)+\cot(\alpha)=a\quad(a\gt 0),$ find $\sqrt{\tan(\alpha)}+\sqrt{\cot(\alpha)}$

Question

I multiplied both expressions: $$\tan\bigl(\sqrt{\tan(\alpha)}+\sqrt{\cot(\alpha)}\bigr)+\cot\bigl(\sqrt{\tan(\alpha)}+\sqrt{\cot(\alpha)}\bigr)=\sqrt{\tan(\alpha)}+\sqrt{\cot(\alpha)}\bigl(a\bigr).$$Raising to the second power also didn't help. The variants are

A)$\sqrt{a+2}$ B)$a-2$ C)$\sqrt{2}+\sqrt{a}$ D)$a+2$ E)$\sqrt{a}-\sqrt{2}$

Answer 1

If $\tan\alpha<0$ then $\cot\alpha<0$, which is contradiction because $a>0$.

Thus, $\tan\alpha>0$, $\cot\alpha>0$ and

$$\sqrt{\tan\alpha}+\sqrt{\cot\alpha}=\sqrt{\tan\alpha+\cot\alpha+2\sqrt{\tan\alpha\cot\alpha}}=\sqrt{a+2}$$