AnswersPrecalculusPractice TestQuestion 8
8
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Practice Test

Question 8 • Precalculus

Melanie simplified the expression the fraction with numerator cotangent of x and denominator open paren 1 over secant of x close paren as shown. Step 1: the fraction with numerator open paren cosine of x over sine of x close paren and denominator cosine of x Step 2: cosine of x over sine of x times 1 over cosine of x Step 3: 1 over sine of x Step 4: tangent of x In which step did Melanie make the first error, and which expression should she have written in that step?

Answer
A
step 4; cosine of x
B
step 1; the fraction with numerator open paren sine of x over cosine of x close paren and denominator cosine of x
C
step 1; the fraction with numerator open paren cosine of x over sine of x close paren and denominator sine of x
D
step 4; cosecant of x

Explanation

Since cotangent of x equals cosine of x over sine of x, and 1 over secant of x equals cosine of x, Melanie’s first three steps are correct. The expression in Step 3, 1 over sine of x, is cosecant of x, not tangent of x. So the first error is in Step 4, where she should have written cosecant of x.

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