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.
 A: By Holder's Inequality $(\cos^2 \theta + \sin^2 \theta)(a \sec \theta+b \csc \theta)(a \sec \theta+b \csc \theta) \ge (a^{\frac{2}{3}}+b^{\frac{2}{3}})^3$
from which it follows that $a \sec \theta+b \csc \theta \ge (a^{\frac{2}{3}}+b^{\frac{2}{3}})^{\frac{3}{2}}$
A: let $$f(\theta)=\frac{a}{\cos(\theta)}+\frac{b}{\sin(\theta)}$$ then the first derivative is given by
$$f'(\theta)=a\frac{(
\sin(\theta))}{\cos(\theta)^2}-b\frac{\cos(\theta)}{\sin(\theta)^2}$$
then you must solve the equation $$f'(\theta)=0$$ for $\theta$.
A: $f(\theta)=a\sec\theta+b\csc\theta,\quad 0<\theta<\frac{\pi}{2}$
$f'(\theta) = a\cdot(-1)\cdot(-\sin\theta)\cdot\sec^2\theta+b\cdot(-1)\cdot(\cos\theta)\cdot\csc^2\theta$
$$f'(\theta)= \frac{a\sin\theta}{\cos^2\theta} -\frac{b\cos\theta}{\sin^2\theta}$$
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
A: Another equivalent approach is to consider $x= \cos \theta$ and $y = \cos \theta$ which lie on the unit circle.
This means you have the minimize the function: $$f(x,y) = \frac{a}{x}+\frac{b}{y}$$
with the constraint $$x^2+y^2=1$$
for $x,y >0$. Lagrange multipliers will do the rest.
A: If calculus is not mandatory,
as $a,b,\sec\theta,\csc\theta>0$
$$\dfrac{a\sec\theta+b\csc\theta}2\ge\sqrt{a\sec\theta\cdot b\csc\theta}$$
The equality occurs if $a\sec\theta=b\csc\theta$
$\iff\dfrac a{\cos\theta}=\dfrac b{\sin\theta}=+\dfrac{\sqrt{a^2+b^2}}1$  $'+'$ as $0<\theta<\dfrac\pi2$
$\implies\sec\theta=\dfrac{\sqrt{a^2+b^2}}a$ and $\csc\theta=?$
