Minimum value of a trigonometric function 
Prove that the minimum value of $$a \sec \theta + b \csc \theta$$ is $$(a^{2/3} + b^{2/3})^{3/2}$$ where $0 < \theta < \frac{\pi}{2}$ and $a$ and $b$ are positive real numbers.

This is one part of a larger question whose first part I solved. I tried to express $a \sec \theta + b \csc \theta$ in a more concise manner, but couldn't do so without introducing more variables. 
Even after equating the function to the minimum, it is not clear to me how to find the value of $\theta$ for which the function attains its minimum.

$f(\theta)=a\sec\theta+b\csc\theta,\quad 0<\theta<\frac{\pi}{2}$
$f'(\theta) = a\cdot(-1)\cdot(-\sin\theta)\cdot\sec^2\theta+b\cdot(-1)\cdot(\cos\theta)\cdot\csc^2\theta$
$$f'(\theta)= \frac{a\sin\theta}{\cos^2\theta} -\frac{b\cos\theta}{\sin^2\theta}$$
Setting $f'(\theta)=0:$
$$f'(\theta) = \frac{a\sin^3\theta-b\cos^3\theta}{\sin^2\theta\cos^2\theta}=0$$
$$\implies a\sin^3\theta=b\cos^3\theta \implies \tan^3\theta=\frac{b}{a}\implies \tan\theta=\bigg(\frac{b}{a}\bigg)^{\frac{1}{3}}$$
Then $\cos\theta = \frac{a^{1/3}}{\sqrt{a^{2/3}+b^{2/3}}}\quad$ and $\sin\theta = \frac{b^{1/3}}{\sqrt{a^{2/3}+b^{2/3}}}$
Then $f(\theta) = a\cdot\frac{\sqrt{a^{2/3}+b^{2/3}}}{a^{1/3}}+b\cdot\frac{\sqrt{a^{2/3}+b^{2/3}}}{b^{1/3}} =(a^{2/3}+b^{2/3})\cdot \sqrt{a^{2/3}+b^{2/3}} = (a^{2/3}+b^{2/3})^{3/2}$
I leave it to you to show that the point is a minimum
