Geometry: Common Core (15th Edition)

Published by Prentice Hall
ISBN 10: 0133281159
ISBN 13: 978-0-13328-115-6

Chapter 8 - Right Triangles and Trigonometry - 8-3 Trigonometry - Practice and Problem-Solving Exercises - Page 512: 51

Answer

Let's rewrite the expression on the left in terms of the ratios of sides: sin $A$ = $\frac{opposite}{hypotenuse}$ tan $A$ = $\frac{opposite}{adjacent}$ Let's plug in what we know into the equation we are given: $\frac{1}{\frac{a^2}{c^2}} - \frac{1}{\frac{a^2}{b^2}} = 1$ Rewrite the equation to simplify: $\frac{c^2}{a^2} - \frac{b^2}{a^2} = 1$ Rewrite the fractions as one fraction: $\frac{c^2 - b^2}{a^2} = 1$ We now look at the Pythagorean theorem, which relates the sides to the hypotenuse. The Pythagorean theorem is given by the following formula: $a^2 + b^2 = c^2$ If we were to rewrite the Pythagorean theorem in terms of $a^2$, then we would have: $a^2 = c^2 - b^2$ We can replace $c^2 - b^2$ with $a^2$: $\frac{a^2}{a^2} = 1$ Simplify the fraction by dividing both the numerator and denominator by their greatest common factor: $1 = 1$ The identity is true.

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