Check the solution to "How to simplify sec2x−1sin2x?"

Ross Colon

Answered question

2023-01-09

How to simplify

\frac{\sec^{2} x - 1}{\sin^{2} x}

possetzvjm

Beginner2023-01-10Added 7 answers

Transform each and every trigonometric function first to

\sin (x)

and

\cos (x)

\frac{\sec^{2} (x) - 1}{\sin^{2} (x)}

= \frac{\frac{1}{\cos^{2} (x)} - 1}{\sin^{2} (x)}

= \frac{\frac{1 - \cos^{2} (x)}{\cos^{2} (x)}}{\sin^{2} (x)}

Use the identity

\sin^{2} (x) + \cos^{2} (x) = 1

= \frac{\frac{\sin^{2} (x)}{\cos^{2} (x)}}{\sin^{2} (x)}

Canceling out the

\sin^{2} (x)

present in both the numerator and the denominator:

= \frac{1}{\cos^{2} (x)}

= \sec^{2} (x)

Raphael Stark

Beginner2023-01-11Added 1 answers

Knowing that,

\sec^{2} x - 1 = \tan^{2} x

Hence,

\frac{\sec^{2} x - 1}{\sin^{2} x}

\frac{\tan^{2} x}{\sin^{2} x}

\frac{\sin^{2} x}{\cos^{2} x} \cdot \frac{1}{\sin^{2} x}

\frac{1}{\cos^{2} x}

\sec^{2} x

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