Problem 54 - Precalculus Essentials

Question

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

Step-by-Step Solution

Answer

\(\frac{\sin \theta}{1-\cot \theta}-\frac{\cos \theta}{\tan \theta-1} = \sin \theta +\cos \theta\). The identity has been verified.

Step 1: Simplify using trigonometric identities

First, it is important to remember that \(\cot \theta = \frac{1}{\tan \theta}\) and \(\tan \theta = \frac{\sin \theta}{\cos \theta}\). Using these identities, the left-hand side of the equation could be rewritten as: \[\frac{\sin \theta}{1-\frac{1}{\tan \theta}}-\frac{\cos \theta}{\frac{\sin \theta}{\cos \theta}-1}\]

Step 2: Use common denominators to simplify

In order to simplify the expression, calculate the common denominators: \[\frac{\sin \theta \tan \theta}{\tan \theta-1}-\frac{\cos^2 \theta}{\sin \theta - \cos \theta}\]

Step 3: Convert terms to same trigonometric function

Rewrite the tangent in the first term in terms of sine and cosine:\[\frac{\sin^2 \theta}{\sin \theta - \cos \theta} - \frac{ \cos^2 \theta}{\sin \theta - \cos \theta}\]

Step 4: Combine under one fraction

Combine the two fractions into one:\[\frac{\sin^2 \theta - \cos^2 \theta}{\sin \theta - \cos \theta}\].

Step 5: Final Step: Use Identity for Difference of Squares

At this stage, the identity \(a^2 - b^2 = (a+b)(a-b)\) can be applied to the numerator to simplify it:\[\frac{(\sin \theta + \cos \theta)(\sin \theta - \cos \theta)}{\sin \theta - \cos \theta}\]. This simplifies to \(\sin \theta + \cos \theta\) which is the right-hand side of the original equation.