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**Unformatted text preview: **nd the values of the circular functions of an angle and solve equations
and inequalities. In Calculus, they are needed to simplify otherwise complicated expressions. In
the next example, we make good use of the Theorems 10.6 and 10.8.
Example 10.3.3. Verify the following identities. Assume that all quantities are deﬁned.
1. 1
= sin(θ)
csc(θ) 4. sec(θ)
1
=
1 − tan(θ)
cos(θ) − sin(θ) 2. tan(θ) = sin(θ) sec(θ) 5. 6 sec(θ) tan(θ) = 3. (sec(θ) − tan(θ))(sec(θ) + tan(θ)) = 1 6. 3
3
−
1 − sin(θ) 1 + sin(θ) sin(θ)
1 + cos(θ)
=
1 − cos(θ)
sin(θ) Solution. In verifying identities, we typically start with the more complicated side of the equation
and use known identities to transform it into the other side of the equation.
1. To verify 1
csc(θ) = sin(θ), we start with the left side. Using csc(θ) =
1
=
csc(θ) 1
sin(θ) , we get: 1
= sin(θ),
1
sin(θ) which is what we were trying to prove.
2. Starting with the right hand side of tan(θ) = sin(θ) sec(θ), we use sec(θ) =
sin(θ) sec(θ) = sin(θ) 1
cos(θ) and ﬁnd: 1
sin(θ)
=
= tan(θ),
cos(θ)
cos(θ) where the last equality is courtesy of Theorem 10.6.
3. Expanding the...

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