Problem 59 - Precalculus Essentials

Question

Verify each identity. $$\frac{\cos ^{2} x-\sin ^{2} x}{1-\tan ^{2} x}=\cos ^{2} x$$

Step-by-Step Solution

Answer

After rewriting the denominator using the Pythagorean identity and simplifying the left side of the equation, we find that the left and right sides are equal - the given identity is thus verified.

Step 1: Understanding the Pythagorean identity

The Pythagorean identity in trigonometry is \( \sin^2x + \cos^2x = 1\). This will be a crucial tool in this proof. Our goal is to eventually manipulate the left side of the original equation to match the right side, or vice versa.

Step 2: Rewrite the denominator

Let's rewrite the denominator of the left side of the given equation, using the Pythagorean identity. Recall that \( \tan^2x = \frac{\sin^2x}{\cos^2x}\). Rewrite \(1 - \tan^2x\) as \( \frac{\cos^2x - \sin^2x}{\cos^2x}\).\nThe equation now reads: \( \frac{\cos^2x - \sin^2x}{\frac{\cos^2x - \sin^2x}{\cos^2x}} = \cos^2x\)

Step 3: Simplify left side of the equation

Dividing by a fraction is the same as multiplying by its reciprocal. So, rewrite the fraction as multiplication: \(\frac{\cos^2x - \sin^2x}{\cos^2x}\)

Step 4: Further Simplify left side of the equation

The numerator and denominator on the left side of the equation both have a factor of \( \cos^2x \). Cancel out these factors to get: \( \cos^2x \)

Step 5: Compare both sides of the equation

By observing the equation, after the simplifications, we see that the left side of the equation is now equal to the right side of the equation, verifying the identity. \( \cos^2x = \cos^2x \)