QuestionProve that $\frac{1-\cos ^{2} x}{\sec ^{2} x-1} = 1-\sin ^{2} x$ .

Studdy Solution

STEP 1

Assumptions1. We are given the trigonometric identity $\frac{1-\cos ^{} x}{\sec ^{} x-1} \equiv1-\sin ^{} x$ . . We know the Pythagorean identity $\sin^x + \cos^x =1$ .
3. We know the reciprocal identity $\sec x = \frac{1}{\cos x}$ .

STEP 2

We begin by simplifying the left side of the equation. We can rewrite $\sec^2x$ as $\frac{1}{\cos^2x}$ .
$\frac{1-\cos ^{2} x}{\frac{1}{\cos^2x}-1}$

STEP 3

Next, we multiply the numerator and the denominator of the fraction by $\cos^2x$ to get rid of the fraction in the denominator.
$\frac{(1-\cos ^{2} x)\cos^2x}{1-\cos^2x}$

STEP 4

implify the expression. The $(1-\cos^2x)$ terms cancel out in the numerator and the denominator.
$\cos^2x$

STEP 5

We know from the Pythagorean identity that $\cos^2x =1 - \sin^2x$ .
$1 - \sin^2x$ This is the same as the right side of the original equation, so the identity is verified.
$1 - \sin^2x \equiv1 - \sin^2x$

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