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**Hint:**First, the question seems to be wrong. In the given equation, put $x = \dfrac{\pi }{4}$ , we get the left-hand side equal to ${\sec ^2}\dfrac{\pi }{4}\csc \dfrac{\pi }{4} = 2\sqrt 2 $ and the right-hand side equal to ${\sec ^2}\dfrac{\pi }{4} + {\csc ^2}\dfrac{\pi }{4} = 2 + 2 = 4$ . Clearly, the left-hand side is not equal to the right-hand side. The correct equation should be ${\sec ^2}x{\csc ^2}x = {\sec ^2}x + {\csc ^2}x$ . Now to establish this identity we have to show that the part on the right side of the equal to sign is equal to the part on the left side of the equal to sign. For that, we will take any one side and solve it using the trigonometric ratios or identities and make it equal to the other side.

**Complete step by step solution:**

We have to prove that ${\sec ^2}x{\csc ^2}x = {\sec ^2} + {\csc ^2}x$

We know that $\sec x = \dfrac{1}{{\cos x}}$ and $\csc x = \dfrac{1}{{\sin x}}$

Solving the right-hand side –

$

{\sec ^2}x + {\csc ^2}x = \dfrac{1}{{{{\cos }^2}x}} + \dfrac{1}{{{{\sin }^2}x}} \\

\Rightarrow {\sec ^2}x + {\csc ^2}x = \dfrac{{{{\sin }^2}x + {{\cos }^2}x}}{{{{\sin }^2}x{{\cos }^2}x}} \\

$

We know that ${\sin ^2}x + {\cos ^2}x = 1$ , so we get –

$

\Rightarrow {\sec ^2}x + {\csc ^2}x = \dfrac{1}{{{{\sin }^2}x{{\cos }^2}x}} \\

\Rightarrow {\sec ^2}x + {\csc ^2}x = {\sec ^2}x{\csc ^2}x \\

$

Thus, the right-hand side comes out to be equal to the left-hand side.

Hence proved that ${\sec ^2}x{\csc ^2}x = {\sec ^2}x + {\csc ^2}x$ .

**Note:**The two sides of a right-angled triangle and one angle other than the right angle are interrelated with each other by trigonometric ratios. The sine of an angle is equal to the ratio of perpendicular and the hypotenuse of the right-angled triangle and the cosine of an angle is equal to the ratio of the base and the hypotenuse of the right-angled triangle. Secant is the reciprocal of the cosine function and cosecant is the reciprocal of the sine function. Identities are used to generalize a relation so that it can be used in other big problems to solve them easily.

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