Exam-Style Problem
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Problem 18
18
☆
Prove the identity
\(\displaystyle \frac{1-\cos^2x}{\sec^2x-1}\equiv 1-\sin^2x.\)
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Solution
\[ \frac{1-\cos^2x}{\sec^2x-1}=\frac{\sin^2x}{\tan^2x} =\frac{\sin^2x}{\frac{\sin^2x}{\cos^2x}}=\cos^2x=1-\sin^2x. \]
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