Problem 42 - Precalculus Essentials

Question

Verify each identity. $$\frac{\tan 2 \theta+\cot 2 \theta}{\sec 2 \theta}=\csc 2 \theta$$

Step-by-Step Solution

Answer

The given identity \(\frac{\tan 2 \theta+\cot 2 \theta}{\sec 2 \theta}=\csc 2 \theta\) holds true when broken down and simplified through several steps of trigonometric transformations.

Step 1: Transform all terms into sine and cosine

Transform each trigonometric term into its sine and cosine equivalent. This gives us: \(\frac{\frac{\sin 2 \theta}{\cos 2 \theta}+\frac{\cos 2 \theta}{\sin 2 \theta}}{\frac{1}{\cos 2 \theta}}\)

Step 2: Simplify the expression

Combining like fractions into one and simplifying gives: \( \frac{\sin^2 2 \theta+\cos^2 2 \theta}{\cos 2 \theta*\sin 2 \theta}\). From here we know that \(\sin^2 \theta + \cos^2 \theta = 1\), therefore the term in the numerator simplifies to \( \frac{1}{\cos 2 \theta*\sin 2 \theta}\)

Step 3: Use the reciprocal identities

By trigonometric identities, we know that \(\frac{1}{\sin \theta} = \csc \theta\) and \(\frac{1}{\cos \theta} = \sec \theta\). So, we can transform the above expression into: \( \csc 2 \theta*\sec 2 \theta\)

Step 4: Use the identity \(\sec \theta*\csc \theta = \csc 2 \theta\)

By trigonometric identities, we know that \(\sec \theta*\csc \theta\) is equal to \(\csc 2 \theta\).\ So the expression becomes: \(\csc 2 \theta\). As the output of step 4 matches the right side of the given identity, we have verified the identity