How do you simplify $\left( {1 - {{\cos }^2}\theta } \right)\left( {1 + {{\cot }^2}\theta } \right)$?

Hint: This problem deals with solving the given equation with trigonometric identities and compound sum angles of trigonometric functions. The trigonometric identities are those with which the whole trigonometry is governed. Here basic trigonometric identities are used which are given below:
$ \Rightarrow {\sin ^2}\theta + {\cos ^2}\theta = 1$
From here $1 - {\cos ^2}\theta = {\sin ^2}\theta $
$ \Rightarrow \cos e{c^2}\theta - {\cot ^2}\theta = 1$
From here $1 + {\cot ^2}\theta = {\cos ec^2}\theta $

Complete step-by-step answer:
The given expression is a trigonometric functional expression which is given by: $\left( {1 - {{\cos }^2}\theta } \right)\left( {1 + {{\cot }^2}\theta } \right)$
Considering the expression, as given below:
We know that from basic trigonometric identity that the sum of the squares of the trigonometric sine and cosine ratios are equal to 1, which is expressed below:
$ \Rightarrow {\sin ^2}\theta + {\cos ^2}\theta = 1$
$\therefore 1 - {\cos ^2}\theta = {\sin ^2}\theta $
Similarly applying and using another basic trigonometric identity that the difference of the squares of the trigonometric cosecant and secant ratios are equal to 1, which is expressed below:
$ \Rightarrow \cos e{c^2}\theta - {\cot ^2}\theta = 1$
$\therefore 1 + {\cot ^2}\theta ={\cos ec^2}\theta $
Now substituting these obtained trigonometric expressions, in the given expression, as shown below:
$ \Rightarrow \left( {1 - {{\cos }^2}\theta } \right)\left( {1 + {{\cot }^2}\theta } \right)$
$ \Rightarrow \left( {{{\sin }^2}\theta } \right)\left( {{\cos ec^2}\theta } \right)$
We know that the reciprocal of the sine trigonometric ratio is equal to the cosecant trigonometric ratio, as shown below:
$ \Rightarrow \cos ec\theta = \dfrac{1}{{\sin \theta }}$
Now squaring on both sides, as given below:
$ \Rightarrow{\cos ec^2}\theta = \dfrac{1}{{{{\sin }^2}\theta }}$
Now substituting this expression in the obtained expression $\left( {{{\sin }^2}\theta } \right)\left( {\cos e{c^2}\theta } \right)$, as shown below:
$ \Rightarrow \left( {{{\sin }^2}\theta } \right)\left( {{\cos ec^2}\theta } \right)$
$ \Rightarrow \left( {{{\sin }^2}\theta } \right)\left( {\dfrac{1}{{{{\sin }^2}\theta }}} \right)$
Here the numerator and the denominator gets cancelled as they are equal, as shown below:
$ \Rightarrow \left( {{{\sin }^2}\theta } \right)\left( {\dfrac{1}{{{{\sin }^2}\theta }}} \right) = 1$
Thus the value of the expression $\left( {1 - {{\cos }^2}\theta } \right)\left( {1 + {{\cot }^2}\theta } \right)$ is equal to 1.

$\therefore \left( {1 - {{\cos }^2}\theta } \right)\left( {1 + {{\cot }^2}\theta } \right) = 1$

Note:
Please note that the formulas of basic trigonometric identities are used to solve this problem. But there are a few other basic trigonometric identities formulas of sine, cosine and tangent, here the basic trigonometric identity formulas for sine, cosine and tangent are given below:
$ \Rightarrow {\sin ^2}\theta + {\cos ^2}\theta = 1$
$ \Rightarrow {\sec ^2}\theta - {\tan ^2}\theta = 1$
$ \Rightarrow {\cos ec^2}\theta - {\cot ^2}\theta = 1$