Stitz-Zeager_College_Algebra_e-book

Turning our attention now to the function g in

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Unformatted text preview: left hand side of the equation gives: (sec(θ) − tan(θ))(sec(θ) + tan(θ)) = sec2 (θ) − tan2 (θ). According to Theorem 10.8, sec2 (θ) = 1 + tan2 (θ). Putting it all together, (sec(θ) − tan(θ))(sec(θ) + tan(θ)) = sec2 (θ) − tan2 (θ) = 1 + tan2 (θ) − tan2 (θ) = 1. 10.3 The Six Circular Functions and Fundamental Identities 641 4. While both sides of our last identity contain fractions, the left side affords us more opportusin( 1 nities to use our identities.4 Substituting sec(θ) = cos(θ) and tan(θ) = cos(θ) , we get: θ) sec(θ) 1 − tan(θ) = = = = = 1 cos(θ) sin(θ) 1− cos(θ) 1 cos(θ) cos(θ) · sin(θ) cos(θ) 1− cos(θ) 1 (cos(θ)) cos(θ) sin(θ) 1− (cos(θ)) cos(θ) 1 sin(θ) (1)(cos(θ)) − (cos(θ)) cos(θ) 1 , cos(θ) − sin(θ) which is exactly what we had set out to show. 5. The right hand side of the equation seems to hold more promise. We get common denominators and add: 3 3 − 1 − sin(θ) 1 + sin(θ) = 3(1 + sin(θ)) 3(1 − sin(θ)) −...
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