Problem 44 - Precalculus Essentials

Question

Verify each identity. $$\frac{\cot x+\cot y}{1-\cot x \cot y}=\frac{\cos x \sin y+\sin x \cos y}{\sin x \sin y-\cos x \cos y}$$

Step-by-Step Solution

Answer

The given trigonometric identity has been successfully verified through the process of translating the cotangent terms into equivalent sine and cosine terms and simplifying the resulting equation.

Step 1: Expanding the identities

We start by converting the \(\cot\) terms to their equivalent \(\cos/\sin\) terms:$$\frac{\cot x+\cot y}{1-\cot x \cot y}=\frac{\frac{\cos x }{\sin x }+\frac{\cos y }{\sin y}}{1-\frac{\cos x }{\sin x }*\frac{\cos y }{\sin y}}.$$

Step 2: Simplify the expression

Next, we find the common denominator to simplify the fractions:$$=\frac{\sin y \cos x + \sin x \cos y}{\sin x \sin y-\cos x \cos y}$$. The numerator and the denominator of this fraction match the ones from the RHS of the original equation.

Step 3: Verification

As both sides of the equation are now the same, we have successfully verified the identity: $$\frac{\cot x+\cot y}{1-\cot x \cot y}=\frac{\cos x \sin y+\sin x \cosy}{\sin x \sin y-\cos x \cos y}$$.