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QuestionMatch the expressions using Quotient and Reciprocal Identities:
1. −∨1csc⁡(θ)-\vee \frac{1}{\csc (\theta)}
2. −v1tan⁡(θ)-v \frac{1}{\tan (\theta)}
3. −vsin⁡(θ)cos⁡(θ)-v \frac{\sin (\theta)}{\cos (\theta)}
4. −v1sec⁡(θ)-v \frac{1}{\sec (\theta)}
5. −∨1sin⁡(θ)-\vee \frac{1}{\sin (\theta)}
6. −v1cot⁡(θ)-v \frac{1}{\cot (\theta)}
7. −vcos⁡(θ)sin⁡(θ)-v \frac{\cos (\theta)}{\sin (\theta)}
8. −v1cos⁡(θ)-v \frac{1}{\cos (\theta)}
Options: a. sin⁡(θ)\sin (\theta), b. cos⁡(θ)\cos (\theta), c. tan⁡(θ)\tan (\theta), d. csc⁡(θ)\csc (\theta), e. sec⁡(θ)\sec (\theta), f. cot⁡(θ)\cot (\theta).

Studdy Solution

STEP 1

Assumptions1. We are given a set of trigonometric expressions. . We need to use the Quotient and Reciprocal Identities to match equivalent expressions.
3. The Quotient and Reciprocal Identities are as followsQuotient Identitiestan⁡(θ)=sin⁡(θ)cos⁡(θ)\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}cot⁡(θ)=cos⁡(θ)sin⁡(θ)\cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)}Reciprocal Identitiescsc⁡(θ)=1sin⁡(θ)\csc(\theta) = \frac{1}{\sin(\theta)}sec⁡(θ)=1cos⁡(θ)\sec(\theta) = \frac{1}{\cos(\theta)}cot⁡(θ)=1tan⁡(θ)\cot(\theta) = \frac{1}{\tan(\theta)}tan⁡(θ)=1cot⁡(θ)\tan(\theta) = \frac{1}{\cot(\theta)}

STEP 2

Let's start by simplifying each expression using the Quotient and Reciprocal Identities.
The first expression is −∨1csc⁡(θ)-\vee \frac{1}{\csc (\theta)}. Using the reciprocal identity for csc⁡(θ)\csc(\theta), we can rewrite this as−∨1csc⁡(θ)=−∨sin⁡(θ)-\vee \frac{1}{\csc (\theta)} = -\vee \sin(\theta)

STEP 3

The second expression is −v1tan⁡(θ)-v \frac{1}{\tan (\theta)}. Using the reciprocal identity for tan⁡(θ)\tan(\theta), we can rewrite this as−v1tan⁡(θ)=−vcot⁡(θ)-v \frac{1}{\tan (\theta)} = -v \cot(\theta)

STEP 4

The third expression is −vsin⁡(θ)cos⁡(θ)-v \frac{\sin (\theta)}{\cos (\theta)}. Using the quotient identity for tan⁡(θ)\tan(\theta), we can rewrite this as−vsin⁡(θ)cos⁡(θ)=−vtan⁡(θ)-v \frac{\sin (\theta)}{\cos (\theta)} = -v \tan(\theta)

STEP 5

The fourth expression is −v1sec⁡(θ)-v \frac{1}{\sec (\theta)}. Using the reciprocal identity for sec⁡(θ)\sec(\theta), we can rewrite this as−v1sec⁡(θ)=−vcos⁡(θ)-v \frac{1}{\sec (\theta)} = -v \cos(\theta)

STEP 6

The fifth expression is −∨1sin⁡(θ)-\vee \frac{1}{\sin (\theta)}. Using the reciprocal identity for sin⁡(θ)\sin(\theta), we can rewrite this as−∨1sin⁡(θ)=−∨csc⁡(θ)-\vee \frac{1}{\sin (\theta)} = -\vee \csc(\theta)

STEP 7

The sixth expression is −v1cot⁡(θ)-v \frac{1}{\cot (\theta)}. Using the reciprocal identity for cot⁡(θ)\cot(\theta), we can rewrite this as−v1cot⁡(θ)=−vtan⁡(θ)-v \frac{1}{\cot (\theta)} = -v \tan(\theta)

STEP 8

The seventh expression is −vcos⁡(θ)sin⁡(θ)-v \frac{\cos (\theta)}{\sin (\theta)}. Using the quotient identity for cot⁡(θ)\cot(\theta), we can rewrite this as−vcos⁡(θ)sin⁡(θ)=−vcot⁡(θ)-v \frac{\cos (\theta)}{\sin (\theta)} = -v \cot(\theta)

STEP 9

The eighth expression is −vcos⁡(θ)-v \frac{}{\cos (\theta)}. Using the reciprocal identity for cos⁡(θ)\cos(\theta), we can rewrite this as−vcos⁡(θ)=−vsec⁡(θ)-v \frac{}{\cos (\theta)} = -v \sec(\theta)

STEP 10

Now, we can match each simplified expression to the corresponding trigonometric function.
. −∨sin⁡(θ)-\vee \sin(\theta) matches with a. sin⁡(θ)\sin (\theta)2. −vcot⁡(θ)-v \cot(\theta) matches with f. cot⁡(θ)\cot (\theta)3. −vtan⁡(θ)-v \tan(\theta) matches with c. tan⁡(θ)\tan (\theta)4. −vcos⁡(θ)-v \cos(\theta) matches with b. cos⁡(θ)\cos (\theta)5. −∨csc⁡(θ)-\vee \csc(\theta) matches with d. csc⁡(θ)\csc (\theta)6. −vtan⁡(θ)-v \tan(\theta) matches with c. tan⁡(θ)\tan (\theta)7. −vcot⁡(θ)-v \cot(\theta) matches with f. cot⁡(θ)\cot (\theta)8. −vsec⁡(θ)-v \sec(\theta) matches with e. sec⁡(θ)\sec (\theta)

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