Simplify the following trigonometric expression:
 $\sec \theta \left( 1-\sin \theta \right)\left( \sec \theta +\tan \theta \right)\times 5$.

Hint: Multiply $\sec \theta $ with $\left( 1-\sin \theta \right)$ and leave $\sec \theta $ as it is when multiplied with 1. Change $\sec \theta $ into $\dfrac{1}{\cos \theta }$ when it is multiplied with $\sin \theta $. Now, change $\dfrac{\sin \theta }{\cos \theta }$ in $\tan \theta $ and then multiply this obtained expression with $\left( \sec \theta +\tan \theta \right)$. Use the trigonometric identity given by ${{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1$ to get the simplified result.

Complete step-by-step answer:
In mathematics, trigonometric identities are equalities that involve trigonometric functions and are true for every value of the occurring variables where both sides of the equality are defined. Geometrically, these are identities involving certain functions of one or more angles. They are distinct from triangle identities, which are identities involving angles and side lengths of a triangle.
Now let us come to the question. We have been given, $\sec \theta \left( 1-\sin \theta \right)\left( \sec \theta +\tan \theta \right)\times 5$. Let the value of this expression is $E$. Therefore,
$E=\sec \theta \left( 1-\sin \theta \right)\left( \sec \theta +\tan \theta \right)\times 5$
This can be written as,
$E=\left( \sec \theta -\sec \theta \sin \theta \right)\left( \sec \theta +\tan \theta \right)\times 5$
Writing, $\sec \theta =\dfrac{1}{\cos \theta }$, we get,
$E=\left( \sec \theta -\dfrac{\sin \theta }{\cos \theta } \right)\left( \sec \theta +\tan \theta \right)\times 5$
We know that, $\dfrac{\sin \theta }{\cos \theta }=\tan \theta $. Therefore,
$E=\left( \sec \theta -\tan \theta \right)\left( \sec \theta +\tan \theta \right)\times 5$
Using the algebraic identity: $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$, we get,
$E=\left( {{\sec }^{2}}\theta -{{\tan }^{2}}\theta \right)\times 5$
Now, using the trigonometric identity: ${{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1$, we get,
$E=1\times 5=5$

Note: We can also simplify the given expression by changing $\sec \theta $ into $\dfrac{1}{\cos \theta }$ and $\tan \theta $ into $\dfrac{\sin \theta }{\cos \theta }$ in the whole expression. We will multiply the numerator and denominator of the two terms and then use the identity, $1-{{\sin }^{2}}\theta ={{\cos }^{2}}\theta $, so that the numerator gets cancelled by the denominator and we will get the simplified value.