Problem 41 - Precalculus Essentials

Question

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

Step-by-Step Solution

Answer

After using trigonometric identities to simplify the given equation, we find that indeed, \(\frac{\tan 2 \theta+\cot 2 \theta}{\csc 2 \theta} = \sec 2 \theta\).

Step 1: Expand using Trigonometric Identities

Expand the equations using the basic trigonometric identities. Write \(\tan 2 \theta\) as \(\frac{\sin 2 \theta}{\cos 2 \theta}\), \(\cot 2 \theta\) as \(\frac{\cos 2 \theta}{\sin 2 \theta}\), \(\csc 2 \theta\) as \(\frac{1}{\sin 2 \theta}\), and \(\sec 2 \theta\) as \(\frac{1}{\cos 2 \theta}\). We get: \(\frac{\frac{\sin 2 \theta}{\cos 2 \theta} +\frac{\cos 2 \theta}{\sin 2 \theta}}{\frac{1}{\sin 2 \theta}}\).

Step 2: Simplify the Complex Fraction

Now, let's simplify this complex fraction. To do this, multiply the numerator and the denominator by \(\sin 2 \theta\cos 2 \theta\) (the LCD of the complex fraction). The equation simplifies to: \(\frac{\sin^2 2 \theta+\cos^2 2 \theta}{\sin 2 \theta}\).

Step 3: Apply Pythagorean Identity

The top of the fraction, \(\sin^2 2 \theta+\cos^2 2 \theta\), is the Pythagorean Identity and equals 1, as per the Pythagorean identity. So we are left with \(\frac{1}{\sin 2 \theta}\).

Step 4: Demonstrate Equality

Our final equation \(\frac{1}{\sin 2 \theta}\) is equivalent to our original right hand side \(\sec 2 \theta\), which is \(\frac{1}{\cos 2 \theta}\). Thus, our original equation \(\frac{\tan 2 \theta+\cot 2 \theta}{\csc 2 \theta} = \sec 2 \theta\) has been verified.