McGraw Hill Glencoe Algebra 2, 2012
MH
McGraw Hill Glencoe Algebra 2, 2012 View details
Mid-Chapter Quiz
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Exercise 17 Page 892

Start on the left-hand side of the equation and use the Pythagorean, Reciprocal, and Quotient Identities to arrive at the right-hand side.

See solution.

Practice makes perfect

We want to verify the given trigonometric identity. sin θ * sec θ/sec θ - 1=(sec θ + 1) cot θ We will start on the left-hand side and use the Pythagorean, Reciprocal, and Quotient Identities to arrive at the right-hand side. Let's begin by performing some basic operations on the left-hand side.

sin θ * sec θ/sec θ - 1
sin θ * sec θ (sec θ + 1)/(sec θ - 1)(sec θ + 1)
sin θ * sec θ (sec θ + 1)/sec ^2 θ - 1^2
sin θ * sec θ (sec θ + 1)/sec ^2 θ - 1

Now let's recall one of the Pythagorean Identities.

tan ^2 θ + 1 = sec ^2 θ We will substitute tan ^2 θ + 1 for sec ^2 θ in our expression. Then we will continue simplifying our expression.

sin θ * sec θ (sec θ + 1)/sec ^2 θ - 1
sin θ * sec θ (sec θ + 1)/tan ^2 θ + 1 - 1
sin θ * sec θ (sec θ + 1)/tan ^2 θ

Next, we will recall one of the Reciprocal Identities. sec θ = 1/cos θ, cos θ ≠ 0 We can substitute 1cos θ for sec θ and simplify our expression.

sin θ * sec θ (sec θ + 1)/tan ^2 θ
sin θ * 1cos θ * (sec θ + 1)/tan ^2 θ
sin θcos θ * (sec θ + 1)/tan ^2 θ

Now let's recall one of the Quotient Identities. tan θ = sin θ/cos θ, cos θ ≠ 0 We can substitute tan θ for sin θcos θ and simplify our expression.

sin θcos θ * (sec θ + 1)/tan ^2 θ
tan θ (sec θ + 1)/tan ^2 θ
â–¼
Simplify
tan θ (sec θ + 1)/tan θ tan θ
sec θ + 1/tan θ

Finally, we will recall once again one of the Reciprocal Identities. tan θ = 1/cot θ, cot θ ≠ 0 We can substitute 1cot θ for tan θ and simplify our expression.

sec θ + 1/tan θ
sec θ + 1/1cot θ
â–¼
Simplify
(sec θ + 1)cot θ/1
(sec θ + 1) cot θ ✓

We started on the left-hand side of the identity and used the Pythagorean, Reciprocal, and Quotient Identities to arrive at the right-hand side. Therefore, we have verified the identity.