How do you prove \[{{\csc }^{2}}\left( x \right)-{{\cot }^{2}}\left( x \right)=1\]?

Hint: To solve this question, we will need to use the trigonometric identity which states that \[{{\sin }^{2}}\left( x \right)+{{\cos }^{2}}\left( x \right)=1\]. To prove an equation, we need to take one of the left-hand side (LHS) or right-hand side (RHS). Simplify the side and express it as the other side.

Complete step by step answer:
We are asked to prove \[{{\csc }^{2}}\left( x \right)-{{\cot }^{2}}\left( x \right)=1\]. Here the left-hand side is \[{{\csc }^{2}}\left( x \right)-{{\cot }^{2}}\left( x \right)\], and the right-hand side is \[1\]. We need to choose one of the sides and simplify the side so it can be expressed as the other side.
Let’s choose the left-hand side \[{{\csc }^{2}}\left( x \right)-{{\cot }^{2}}\left( x \right)\], we know that \[\csc x=\dfrac{1}{\sin x}\], and \[\cot x=\dfrac{\cos x}{\sin x}\]. Using this in the expression, it can be expressed as
\[\Rightarrow {{\left( \dfrac{1}{\sin x} \right)}^{2}}-{{\left( \dfrac{\cos x}{\sin x} \right)}^{2}}\]
\[\Rightarrow \dfrac{1}{{{\sin }^{2}}x}-\dfrac{{{\cos }^{2}}x}{{{\sin }^{2}}x}\]
As the denominator for both terms in the above expression is the same, we can subtract the numerators,
\[\Rightarrow \dfrac{1-{{\cos }^{2}}x}{{{\sin }^{2}}x}\]
Now, we know the trigonometric identity which states that \[{{\sin }^{2}}\left( x \right)+{{\cos }^{2}}\left( x \right)=1\], subtracting \[{{\cos }^{2}}\left( x \right)\] from both sides of this identity, we get
\[\Rightarrow {{\sin }^{2}}\left( x \right)+{{\cos }^{2}}\left( x \right)-{{\cos }^{2}}\left( x \right)=1-{{\cos }^{2}}\left( x \right)\]
\[\Rightarrow {{\sin }^{2}}\left( x \right)=1-{{\cos }^{2}}\left( x \right)\]
Using this identity in the above expression of the left-hand side, we get
\[\Rightarrow \dfrac{1-{{\cos }^{2}}x}{{{\sin }^{2}}x}=\dfrac{1-{{\cos }^{2}}x}{1-{{\cos }^{2}}x}\]
\[\Rightarrow 1\]
After simplifying the left-hand side, we get the 1, as the right-hand side of the given expression is also 1. Hence, we proved that the left-hand side and right-hand sides are equal.
Hence, \[{{\csc }^{2}}\left( x \right)-{{\cot }^{2}}\left( x \right)=1\] is proved.

Note: The expression given in the question is one of the trigonometric identities similar to \[{{\sin }^{2}}\left( x \right)+{{\cos }^{2}}\left( x \right)=1\]. We can also prove that \[1+{{\tan }^{2}}\left( x \right)={{\sec }^{2}}\left( x \right)\] by using a similar method to this question.
The following identities are useful while solving the question on proofs or evaluating expressions, so it should be remembered:
\[{{\sin }^{2}}\left( x \right)+{{\cos }^{2}}\left( x \right)=1\]
\[{{\csc }^{2}}\left( x \right)=1+{{\cot }^{2}}\left( x \right)\]
\[1+{{\tan }^{2}}\left( x \right)={{\sec }^{2}}\left( x \right)\]