Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 6 - Analytic Trigonometry - Section 6.5 Sum and Difference Formulas - 6.5 Assess Your Understanding - Page 509: 67

Answer

See proof below.

Work Step by Step

Since, $sec (A+B)=\dfrac{1}{\cos(A+B)}$ Apply the sum formula : $\cos(A+B)=\cos(A)\cos(B) -\sin(A)\sin(B)$ In order to prove the given identity, we simplify the left hand side $\text{RHS}$ as follows: $\text{LHS}=\dfrac{\csc a \csc B}{\cot a \cot B -1} \\=\dfrac{(\dfrac{1}{\sin a}) \times (\dfrac{1}{\sin B})}{(\dfrac{\cos a }{\sin a} ) \times (\dfrac{\cos B}{\sin B})-1}\\=\dfrac{1}{(\cos a) (\cos B)-(\sin a) (\sin B)}\\=\dfrac{1}{\cos (a+B)}\\=\sec (a+B) \\=\text{ LHS}$
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