Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 6 - Review Exercises - Page 291: 33


$tan(arcsin\frac{3}{5}+arccos\frac{5}{7}) = \frac{588+125\sqrt{24}}{184}$

Work Step by Step

Let $A = arcsin\frac{3}{5}$ Then: $tan~A = \frac{3}{\sqrt{5^2-3^2}} = \frac{3}{4}$ Let $B = arccos(\frac{5}{7})$ Then: $tan~B = \frac{\sqrt{7^2-5^2}}{5} = \frac{\sqrt{24}}{5}$ We can find $~~tan(arcsin\frac{3}{5}+arccos\frac{5}{7})$: $tan(A+B) = \frac{tan~A+tan~B}{1-tan~A~tan~B}$ $tan(A+B) = \frac{\frac{3}{4}+\frac{\sqrt{24}}{5}}{1-(\frac{3}{4})(\frac{\sqrt{24}}{5})}$ $tan(A+B) = \frac{\frac{15+4\sqrt{24}}{20}}{\frac{20}{20}-\frac{3\sqrt{24}}{20}}$ $tan(A+B) = (\frac{15+4\sqrt{24}}{20-3\sqrt{24}})~(\frac{20+3\sqrt{24}}{20+3\sqrt{24}})$ $tan(A+B) = \frac{588+125\sqrt{24}}{184}$ Therefore, $~~tan(arcsin\frac{3}{5}+arccos\frac{5}{7}) = \frac{588+125\sqrt{24}}{184}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.