Minimum value of Complex Trigonometric Expression 
Minimum  value of of $\displaystyle f(\theta) = \frac{a}{\cos \theta}+\frac{b}{\sin \theta}+\sqrt{\frac{a^2}{\cos^2 \theta}+\frac{b^2}{\sin^2 \theta}}.$
Where  $\displaystyle a,b>0, \theta \in \bigg(0,\frac{\pi}{2}\bigg).$

what i try
$$f(\theta)=\frac{2(a\sin \theta +b\cos \theta)+2\sqrt{\bigg(a\sin \theta+b\cos \theta\bigg)^2-a\sin 2 \theta}}{\sin 2 \theta }$$
How do i minimize is Help me please
 A: \begin{align*}
f'(\theta) &= a \tan \theta  \sec \theta + b \tan \theta \sec \theta \hfill \\
&\quad {}+ \frac{2 a^2 \tan \theta \sec^2 \theta + 2 b^2 \tan \theta \sec ^2\theta }{2 \sqrt{a^2 \sec ^2 \theta + b^2 \sec^2 \theta}}  \\
&= \tan \theta \sec \theta \left(\cos \theta \sqrt{\left(a^2+b^2\right) \sec^2 \theta }+a+b\right)  \\
&= \tan \theta \sec \theta \left( a + b + \sqrt{a^2 + b^2} \right)  \text{.}
\end{align*}
$\tan \theta \neq 0$ for $\theta \in (0, \pi/2)$.  $\sec \theta \neq 0$ for any $\theta \in \Bbb{R}$.  Since $a>0, b>0$, the parenthesized expression is always positive.  Therefore, this function has no critical points.
So we check the endpoints to see which one would be the location of the minimum if it were a permissible point.  $f(0) = a + b + \sqrt{a^2 + b^2}$ and $\lim_{\theta \rightarrow \pi/2^-} f(\theta) = \infty$ (requiring the use of $a>0, b>0$).
So for $\theta \in (0,\pi/2)$, $f$ has no minimum value, but it takes values arbitrarily close to $a+b+\sqrt{a^2 + b^2}$, a lower bound for values of $f$, as $\theta$ approaches $0$.
