Prove the following trigonometric expression:
\[\dfrac{{{\operatorname{cosec}}^{2}}\theta -{{\cos }^{2}}\theta }{{{\cot }^{2}}\theta }={{\sec }^{2}}\theta -{{\sin }^{2}}\theta \].

Hint: The given equation is in terms of cosec x and cot x. So, we must use any of the identities including them. To make the equation look simpler, there are many interrelations between cosec x, cot x, cos x, and sin x. As, \[\theta \in R\], we can use the relation of sin x instead of cosec x and sin x, cos x instead of cot x. So, first, convert the whole equation in terms of cos x and sin x. Then, use any of the identities which makes the equation easy to solve. Here, use
\[\cot x=\dfrac{\cos x}{\sin x};\operatorname{cosec}x=\dfrac{1}{\sin x};\sec x=\dfrac{1}{\cos x}\]

Complete step-by-step solution -
Equality with sine, cosine, or tangent in them is called trigonometric equality. These are solved by some interrelations known beforehand. All the interrelations which relate sine, cosine, tangent, secant, cotangent, cosecant are called trigonometric identities. These trigonometric identities solve the equation and make them simpler to understand for proof. These are the main and crucial steps to take us nearer to the result.
Let us consider the LHS part of our question.
\[\dfrac{{{\operatorname{cosec}}^{2}}\theta -{{\cos }^{2}}\theta }{{{\cot }^{2}}\theta }\]
By basic trigonometric knowledge, we know the relation given by:
\[\operatorname{cosec}\theta =\dfrac{1}{\sin \theta }\]
By substituting this in our equation, we get,
\[\dfrac{{{\left( \dfrac{1}{\sin \theta } \right)}^{2}}-{{\cos }^{2}}\theta }{{{\cot }^{2}}\theta }\]
By simplifying the above term in the bracket, we get it as,
\[\dfrac{\dfrac{1}{{{\sin }^{2}}\theta }-{{\cos }^{2}}\theta }{{{\cot }^{2}}\theta }\]
By basic trigonometric knowledge, we know the relation given by:
\[\cot \theta =\dfrac{\cos \theta }{\sin \theta }\]
By substituting this in our equation, we get it as,
\[\dfrac{\dfrac{1}{{{\sin }^{2}}\theta }-{{\cos }^{2}}\theta }{{{\left( \dfrac{\cos \theta }{\sin \theta } \right)}^{2}}}\]
By simplifying the term inside the bracket, we get it as,
\[\dfrac{\dfrac{1}{{{\sin }^{2}}\theta }-{{\cos }^{2}}\theta }{\dfrac{{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta }}\]
By taking the least common multiple in the numerator, we get it as,
\[\dfrac{\dfrac{1-{{\cos }^{2}}\theta {{\sin }^{2}}\theta }{{{\sin }^{2}}\theta }}{\dfrac{{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta }}\]
By canceling the common terms in the fraction, we get it as,
\[\dfrac{1-{{\cos }^{2}}\theta {{\sin }^{2}}\theta }{{{\cos }^{2}}\theta }\]
By separating this fraction into two parts, we get it as,
\[\dfrac{1}{{{\cos }^{2}}\theta }-\dfrac{{{\cos }^{2}}\theta {{\sin }^{2}}\theta }{{{\cos }^{2}}\theta }\]
By canceling the common terms, we get the expression as,
\[\dfrac{1}{{{\cos }^{2}}\theta }-{{\sin }^{2}}\theta \]
By basic trigonometric knowledge, we can say the relation as:
\[\sec \theta =\dfrac{1}{\cos \theta }\]
By substituting this in our equation, we get it as,
\[{{\left( \sec \theta \right)}^{2}}-{{\sin }^{2}}\theta \]
By equating this to the original equation, we get it as,
\[\dfrac{{{\operatorname{cosec}}^{2}}\theta -{{\cos }^{2}}\theta }{{{\cot }^{2}}\theta }={{\sec }^{2}}\theta -{{\sin }^{2}}\theta \]
LHS = RHS
Therefore, we have proved the required equation.

Note: Students have to be careful while substituting cosec x and cot x as they tend to substitute in a reverse way and it leads to wrong answers. Students also need to remember to write \[{{\sin }^{2}}\theta \text{ term at }{{\cos }^{2}}\theta \] before canceling it. Again, the idea of taking LCM and then breaking the fraction into parts is very crucial. Apply them carefully.