McGraw Hill Glencoe Algebra 2, 2012
MH
McGraw Hill Glencoe Algebra 2, 2012 View details
Mid-Chapter Quiz
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Exercise 8 Page 892

Consider using the Pythagorean Identity tan ^2 θ + 1 = sec ^2 θ.

tan θ =sqrt(7)/3

Practice makes perfect

We want to find the exact value of tan θ given that sec θ = 43. To do so, we will use one of the Pythagorean Identities. tan ^2 θ + 1 = sec ^2 θ Let's do it!

sec θ = 4/3
sec ^2 θ = (4/3)^2
sec ^2 θ = 4^2/3^2
sec ^2 θ = 16/9
tan ^2 θ+1 = 16/9
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Solve for tan θ
tan^2 θ = 16/9 - 1
tan^2 θ = 16/9 - 9/9
tan^2 θ = 7/9
tan θ =± sqrt(7/9)
tan θ =± sqrt(7)/sqrt(9)
tan θ =± sqrt(7)/3

Be aware that we are told that θ lies between 0^(∘) and 90^(∘). Therefore, θ is in Quadrant I.

In this quadrant, the sine of θ is positive and the cosine of θ is positive. Since tan θ= sin θcos θ, the sign of tan θ is positive in this quadrant. Therefore, we will only keep the positive solution. tan θ =sqrt(7)/3