Chapter 7.3, Problem 49E

To determine

To verify: that the given equation is an identity.

Answer to Problem 49E

See explanation

Explanation of Solution

Given information:

  ${\sec }^{2}x=\frac{1-{\cos }^{2}x}{1-{\sin }^{2}x}+{\csc }^{2}x-{\cot }^{2}x$

Calculation:

Simplify the right side of the equation into left side as follows:

  $\begin{array}{c}{\sec }^{2}x=\frac{1-{\cos }^{2}x}{1-{\sin }^{2}x}+{\csc }^{2}x-{\cot }^{2}x\end{array}$

  $\begin{array}{c}=\frac{{\sin }^{2}x}{{\cos }^{2}x}+{\csc }^{2}x-{\cot }^{2}x\left[{\text{Pythagorean identiy: sin}}^{2}x+{\cos }^{2}x=1\right]\end{array}$

  $\begin{array}{c}={\tan }^{2}x+{\csc }^{2}x-{\cot }^{2}x\left[\text{Quotient identiy: }\tan x=\frac{\sin x}{\cos x}\right]\end{array}$

  $\begin{array}{c}={\tan }^{2}x+1\left[\text{Pythagorean identiy: }1+{\cot }^{2}x={\csc }^{2}x\right]\end{array}$

  $\begin{array}{c}={\sec }^{2}x\left[\text{Pythagorean identiy: }1+{\tan }^{2}x={\sec }^{2}x\right]\end{array}$

  $\begin{array}{c}LHS=RHS\end{array}$

Since the right side of the equation is transformed into left hand side. So, the equation is an identity.