Question

How do you prove that \[{\sec ^2}x({\csc ^2}x) = {\sec ^2}x + {\csc ^2}x\]?

Answer 1

Hint: Here in this question we have to prove the given inequality which is given in this question. This question involves the trigonometric function we should know about the trigonometry ratio. Hence by using the simple calculations we are going to prove the given inequality.

Complete step-by-step solution:
In the trigonometry we have six trigonometry ratios namely sine, cosine, tangent, cosecant, secant and cotangent. These are abbreviated as sin, cos, tan, csc, sec and cot. The 3 trigonometry ratios are reciprocal of the other trigonometry ratios. Here cosine is the reciprocal of the sine. The secant is the reciprocal of the cosine. The cotangent is the reciprocal of the tangent.
Now consider the given inequality \[{\sec ^2}x({\csc ^2}x) = {\sec ^2}x + {\csc ^2}x\]
Now we consider the RHS \[{\sec ^2}x + {\csc ^2}x\], this can be written in the form of reciprocal of the given trigonometric ratio and this can be written as
\[ \Rightarrow \dfrac{1}{{{{\cos }^2}x}} + \dfrac{1}{{{{\sin }^2}x}}\]
Now take the LCM for the denominator and this can be written as
\[ \Rightarrow \dfrac{{{{\sin }^2}x + {{\cos }^2}x}}{{{{\cos }^2}x.{{\sin }^2}x}}\]
by the trigonometry identity we know that \[{\sin ^2}x + {\cos ^2}x = 1\].Therefore the above inequality is written as
\[ \Rightarrow \dfrac{1}{{{{\cos }^2}x.{{\sin }^2}x}}\]
Since the denominator is in the form of the product, we can write it separately and we write it as
\[ \Rightarrow \dfrac{1}{{{{\cos }^2}x}}.\dfrac{1}{{{{\sin }^2}x}}\]
The 3 trigonometry ratios are reciprocal of the other trigonometry ratios. Here cosine is the reciprocal of the sine. The secant is the reciprocal of the cosine. The cotangent is the reciprocal of the tangent. This can be written as
\[ \Rightarrow {\sec ^2}x.{\csc ^2}x\]
\[ \Rightarrow LHS\]
Here we have proved LHS = RHS.

Note: The question involves the trigonometric functions and we have to prove the trigonometric function. When we simplify the trigonometric functions and which will be equal to the RHS then the function is proved. While simplifying the trigonometric functions we must know about the trigonometric ratios and the trigonometric identities.