How do you prove $\left[ \dfrac{1}{1-\sin x} \right]+\left[ \dfrac{1}{1+\sin x} \right]=2{{\sec }^{2}}x$?

Hint: We first simplify the numerator for the addition part of $\left[ \dfrac{1}{1-\sin x} \right]+\left[ \dfrac{1}{1+\sin x} \right]$ separately. We have to multiply the denominators for the addition to use it as LCM. We use the identities ${{\sin }^{2}}x+{{\cos }^{2}}x=1$. We also know that the terms $\cos x$ and $\sec x$ are inverse of each other. This gives $\dfrac{1}{\cos x}=\sec x$.

Complete step by step answer:
We have been given an addition of trigonometrical fraction of $\left[ \dfrac{1}{1-\sin x} \right]+\left[ \dfrac{1}{1+\sin x} \right]$.
To add them we need to find the LCM of $\left( 1-\sin x \right)$ and $\left( 1+\sin x \right)$ which is their multiplication.
So, $\left[ \dfrac{1}{1-\sin x} \right]+\left[ \dfrac{1}{1+\sin x} \right]=\dfrac{\left( 1+\sin x \right)+\left( 1-\sin x \right)}{\left( 1-\sin x \right)\left( 1+\sin x \right)}$.
We know that $\left( a+b \right)\left( a-b \right)=\left( {{a}^{2}}-{{b}^{2}} \right)$. This is a factorisation identity of square subtraction.
Therefore, the LCM is $\left( 1-\sin x \right)\left( 1+\sin x \right)=\left( 1-{{\sin }^{2}}x \right)$.
The addition on the numerator is $\left( 1+\sin x \right)+\left( 1-\sin x \right)=1+\sin x+1-\sin x=2$.
So, $\left[ \dfrac{1}{1-\sin x} \right]+\left[ \dfrac{1}{1+\sin x} \right]=\dfrac{2}{\left( 1-{{\sin }^{2}}x \right)}$.
For the numerator part we have $1-{{\sin }^{2}}x={{\cos }^{2}}x$ as for any value of $x$, ${{\sin }^{2}}x+{{\cos }^{2}}x=1$.
Therefore, $\left[ \dfrac{1}{1-\sin x} \right]+\left[ \dfrac{1}{1+\sin x} \right]=\dfrac{2}{{{\cos }^{2}}x}$
We know that the terms $\cos x$ and $\sec x$ are inverse of each other. So, $\dfrac{1}{\cos x}=\sec x$.
Squaring we get $\dfrac{1}{{{\cos }^{2}}x}={{\sec }^{2}}x$. We put the value in the equation to get
$\left[ \dfrac{1}{1-\sin x} \right]+\left[ \dfrac{1}{1+\sin x} \right]=2\times \dfrac{1}{{{\cos }^{2}}x}=2{{\sec }^{2}}x$.
Therefore, the simplified solution of $\left[ \dfrac{1}{1-\sin x} \right]+\left[ \dfrac{1}{1+\sin x} \right]$ is $2{{\sec }^{2}}x$. Thus, verified.

Note: We need to remember that the final terms are square terms. The identities ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ and $\dfrac{1}{\cos x}=\sec x$ are valid for any value of $x$. In the case of $\dfrac{1}{\cos x}=\sec x$, the only condition is $\cos x\ne 0$. The division of the fraction part only gives 1 as the solution.