Prove: \[\cos e{c^2}\theta + {\sec ^2}\theta = {\sec ^2}\theta .\cos e{c^2}\theta \]?

Hint: To prove this type of question, we need to use some of the basic trigonometric identities and formulae. To prove the given result, we have to make right side of the equation equals to left side of the equation using trigonometric identities such as $ {\sin ^2}\theta + {\cos ^2}\theta = 1 $ , $ \sec x = \dfrac{1}{{\cos x}} $ and $ \cos ec\left( x \right) = \dfrac{1}{{\sin \left( x \right)}} $ . To prove the desired result, we have to proceed with the left hand side of the equation and simplify and manipulate it to get the right side of the equation.

Complete step-by-step answer:
In the given question, we are required to prove the given trigonometric equation with the help of trigonometric formulae and identities.
So, we proceed with the left side of the equation,
 $ \cos e{c^2}\theta + {\sec ^2}\theta $
Using the basic trigonometric formulae $ \sec x = \dfrac{1}{{\cos x}} $ and $ \cos ec\left( x \right) = \dfrac{1}{{\sin \left( x \right)}} $ , we get,
=\[\dfrac{1}{{{{\sin }^2}\theta }} + \dfrac{1}{{{{\cos }^2}\theta }}\]
Taking LCM of the denominators, we get,
 $ = \dfrac{{{{\cos }^2}\theta + {{\sin }^2}\theta }}{{{{\sin }^2}\theta .{{\cos }^2}\theta }} $
Applying the basic trigonometric identity $ {\sin ^2}\theta + {\cos ^2}\theta = 1 $ , we get,
 $ = \dfrac{1}{{{{\sin }^2}\theta .{{\cos }^2}\theta }} $
Separating the denominators, we get,
 $ = \dfrac{1}{{{{\sin }^2}\theta }}.\dfrac{1}{{{{\cos }^2}\theta }} $
Using the same trigonometric formulae $ \sec x = \dfrac{1}{{\cos x}} $ and $ \cos ec\left( x \right) = \dfrac{1}{{\sin \left( x \right)}} $ again, we get,
 $ = \cos e{c^2}\theta .{\sec ^2}\theta $
Hence, the given trigonometric equation is proved.
So, the correct answer is “ $ \cos e{c^2}\theta .{\sec ^2}\theta $ ”.

Note: Such questions that require us to prove a trigonometric equation require thorough knowledge of trigonometric functions, formulae and trigonometric identities. Algebraic operations and rules such as taking LCM of denominators and transposition of terms to isolate a specific term, play a significant role in the solution of such problems.