Problem 35 - Precalculus Essentials

Question

Verify each identity. $$\frac{\sec x-\csc x}{\sec x+\csc x}=\frac{\tan x-1}{\tan x+1}$$

Step-by-Step Solution

Answer

Upon substitution of trigonometric identities and simplification of both sides of the equation, we get the same result, showing that the identities are indeed equal.

Step 1: Apply the Definitions of Trigonometric Functions

Recall that \(\sec(x) = \frac{1}{\cos(x)}\), \(\csc(x) = \frac{1}{\sin(x)}\), and \(\tan(x) = \frac{\sin(x)}{\cos(x)}\). Applying these substitutions, the given identity transforms into \(\frac{1/\cos(x) - 1/\sin(x)}{1/\cos(x) + 1/\sin(x)} = \frac{\sin(x)/\cos(x) - 1}{\sin(x)/\cos(x) + 1}\).

Step 2: Simplify the Expressions

Find common denominator of \(\cos(x)\) and \(\sin(x)\) for both numerator and denominator of the left side of the equation: \(\frac{\sin(x)-\cos(x)}{\sin(x)+\cos(x)} = \frac{\sin(x)/\cos(x) - 1}{\sin(x)/\cos(x) + 1}\). Similarly, for the right side, find common denominator \(\cos(x)\) for both numerator and denominator to get \(\frac{\sin(x)-\cos(x)}{\sin(x)+\cos(x)}\).

Step 3: Show the Equality

Now, after simplification, both sides of the equation are equal. The identity is proven.