1 = -2 cot?e 1- sec e 1+ sec e To establish the identity, start with the side containing the more complicated expression. Here, expressions on both sides are of the same complexity. Start with the expression on the left side. 1 Add the quotients. 1- sece 1+ sec 0 (1 + sec 0) + (1 – sec 0) (1 – sec 0)(1 + sec 0) (1- sec 0)(1 + sec 0) (1 + sec 0) + (1 – sec 0) (1 + sec 0) – (1 – sec 0) (1 - sec 0)(1 + sec 0) (1 - sec 0)(1+ sec 0) (1+ sec 0) – (1 – sec 0)
1 = -2 cot?e 1- sec e 1+ sec e To establish the identity, start with the side containing the more complicated expression. Here, expressions on both sides are of the same complexity. Start with the expression on the left side. 1 Add the quotients. 1- sece 1+ sec 0 (1 + sec 0) + (1 – sec 0) (1 – sec 0)(1 + sec 0) (1- sec 0)(1 + sec 0) (1 + sec 0) + (1 – sec 0) (1 + sec 0) – (1 – sec 0) (1 - sec 0)(1 + sec 0) (1 - sec 0)(1+ sec 0) (1+ sec 0) – (1 – sec 0)
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter6: The Trigonometric Functions
Section6.2: Trigonometric Functions Of Angles
Problem 65E
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