Question: Prove the inequality:
$$\left(1+\dfrac{1}{\sin a}\right)\left(1+\dfrac{1}{\cos a}\right)\ge 3+2\sqrt{2}; \text{ for } a\in\left]0,\frac{\pi}{2}\right[$$

Answer: 
I think it should be $(0,\frac{\pi}{2})$, so that both $\sin a$ and $\cos a$ are positive and well defined.

$$\bigg (1+\frac{1}{\sin a} \bigg)\bigg (1+\frac{1}{\cos a} \bigg) = 1+ \frac{1}{\sin a} + \frac{1}{\cos a} + \frac{1}{\sin a\cos a}$$

$$\frac{1}{\sin a} + \frac{1}{\cos a} \geq \frac{4}{\sin a + \cos a} \geq \frac{4}{\sqrt{2(\sin^2 a + \cos^2 a)}} = 2\sqrt 2$$

$$\frac{1}{\sin a\cos a} \geq \frac{2}{\sin^2 a + \cos^2 a} = 2$$

So,
$$\bigg (1+\frac{1}{\sin a} \bigg)\bigg (1+\frac{1}{\cos a} \bigg) = 1+ \frac{1}{\sin a} + \frac{1}{\cos a} + \frac{1}{\sin a\cos a} \geq 3+ 2\sqrt 2$$

Equality occurs when $a = \frac{\pi}{4}$
Differentiate to find the minimum of the LHS :

$$\frac{\mathrm{d}}{\mathrm{d}a}\left(\,\left(1+\frac{1}{\sin\,a}\right)\left(1+\frac{1}{\cos\,a}\right)\,\right)=0$$

$$\left(-\frac{1}{\sin^2 a}\right)\cos a\left(1+\frac{1}{\cos a}\right)+\left(1+\frac{1}{\sin a}\right)(-\sin a)\left(-\frac{1}{\cos^2 a}\right)=0$$

$$\left(\sin\,a+1\right)\left(-\frac{1}{\cos^2 a}\right)=\left(\cos\,a+1\right)\left(\frac{1}{\sin^2 a}\right)$$

$$(1+\sin a)(\sin^2a)=(1+\cos a)(\cos^2a)=(1+\cos a)(1-\sin^2a)$$

$$(1+\sin\,a)(\sin^2a)=(1+\cos\,a)(1+\sin\,a)(1-\sin\,a)$$

$$\sin^2a=1-\cos^2a=(1-\cos\,a)(1+\cos\,a)=(1+\cos\,a)(1-\sin\,a)$$

$$1-\cos\,a=1-\sin\,a$$

$$\cos\,a=\sin\,a$$

This only happens at $a=\frac{\pi}{4}$ in the given range.

And here $\sin\,a=\cos\,a=\frac{1}{\sqrt{2}}$.

So the minimum of the LHS is $(1+\sqrt{2})^2=1+2+2\sqrt{2}=3+2\sqrt{2}$