Quotient Identities Proof Quiz (HSF-TF.C.9)

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1) Simplify: sin(x)/cos(x) + cos(x)/sin(x)

1/(sin x cos x)

Tan(x) + cot(x)

2tan(x)

Sec(x) + csc(x)

Step 1: Write over common denominator sinx·cosx: (sin²x + cos²x)/(sinx cosx).

Step 2: Use sin²x + cos²x = 1.

So, the final answer is 1/(sinx cosx).

Explanation

Step 1: Write over common denominator sinx·cosx: (sin²x + cos²x)/(sinx cosx).

Step 2: Use sin²x + cos²x = 1.

So, the final answer is 1/(sinx cosx).

About This Quiz

Quotient Identities Proof Practice Quiz - Quiz

Lock in tan θ = sin θ / cos θ and cot θ = cos θ / sin θ. You’ll prove small equalities (like tan·cos = sin), spot non-identities, and rewrite mixed expressions into one trig function. Clean, justified steps that make later algebra effortless.

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2) If cot(θ) = cos(θ)/sin(θ), what is 1/cot(θ)?

Sec(θ)

Csc(θ)

Cos(θ)

Tan(θ)

Step 1: 1/cotθ = 1 / (cosθ/sinθ) = sinθ/cosθ.

Step 2: sinθ/cosθ = tanθ.

So, the final answer is tanθ.

Explanation

Step 1: 1/cotθ = 1 / (cosθ/sinθ) = sinθ/cosθ.

Step 2: sinθ/cosθ = tanθ.

So, the final answer is tanθ.

3) If tan(A) = 3, then sin(A)/cos(A) equals

1/3

−3

By definition, sinA/cosA = tanA = 3.

Explanation

By definition, sinA/cosA = tanA = 3.

4) Which equals (sin(x) + cos(x))/cos(x)?

Tan(x) + 1

1 + cot(x)

Sec(x) + tan(x)

Csc(x) + 1

Step 1: (sin+cos)/cos = sin/cos + cos/cos = tanx + 1.

So, the final answer is tanx + 1.

Explanation

Step 1: (sin+cos)/cos = sin/cos + cos/cos = tanx + 1.

So, the final answer is tanx + 1.

5) Prove tan(x)·cos(x) = sin(x). Which step correctly applies the quotient identity?

Replace tan(x) with cos(x)/sin(x)

Replace tan(x) with sin(x)/cos(x)

Replace cos(x) with sin(x)/tan(x)

Replace sin(x) with tan(x)·cos(x)

Step 1: tanx·cosx = (sinx/cosx)·cosx.

Step 2: Cancel cosx ⇒ sinx.

So, the final answer is B.

Explanation

Step 1: tanx·cosx = (sinx/cosx)·cosx.

Step 2: Cancel cosx ⇒ sinx.

So, the final answer is B.

6) Which identity proves cot(x)·sin(x) = cos(x)?

Substitute cot(x) = sin(x)/cos(x)

Substitute sin(x) = tan(x)·cos(x)

Substitute cot(x) = cos(x)/sin(x)

Substitute cos(x) = cot(x)·sin(x)

Step 1: cotx·sinx = (cosx/sinx)·sinx.

Step 2: Cancel sinx ⇒ cosx.

So, the final answer is C.

Explanation

Step 1: cotx·sinx = (cosx/sinx)·sinx.

Step 2: Cancel sinx ⇒ cosx.

So, the final answer is C.

7) Simplify: 1/tan(θ)

Sin(θ)

Cos(θ)

Sec(θ)

Cot(θ)

Step 1: 1/tanθ = 1/(sinθ/cosθ) = cosθ/sinθ = cotθ.

So, the final answer is cotθ.

Explanation

Step 1: 1/tanθ = 1/(sinθ/cosθ) = cosθ/sinθ = cotθ.

So, the final answer is cotθ.

8) Evaluate: [cos(x)/sin(x)]·[sin(x)/cos(x)]

Tan²(x)

Sin(x)cos(x)

Step 1: (cos/sin)·(sin/cos) = 1.

So, the final answer is 1.

Explanation

Step 1: (cos/sin)·(sin/cos) = 1.

So, the final answer is 1.

9) Simplify tan(x)/cot(x)

Cot(x)

Tan²(x)

Step 1: tan/cot = (sin/cos)/(cos/sin) = (sin²/cos²) = tan².

So, the final answer is tan²x.

Explanation

Step 1: tan/cot = (sin/cos)/(cos/sin) = (sin²/cos²) = tan².

So, the final answer is tan²x.

10) Simplify: tan(θ)·cot(θ)

Tan²(θ)

Step 1: tanθ·cotθ = (sinθ/cosθ)·(cosθ/sinθ) = 1.

So, the final answer is 1.

Explanation

Step 1: tanθ·cotθ = (sinθ/cosθ)·(cosθ/sinθ) = 1.

So, the final answer is 1.

11) Simplify: cot(x)/csc(x)

Sin(x)

Tan(x)

Cos(x)

Sec(x)

Step 1: cot/csc = (cos/sin)/(1/sin) = (cos/sin)·sin = cos.

So, the final answer is cos(x).

Explanation

Step 1: cot/csc = (cos/sin)/(1/sin) = (cos/sin)·sin = cos.

So, the final answer is cos(x).

12) Simplify sin(x)/tan(x)

Cos(x)

Sin(x)

Tan(x)

Step 1: sin/tan = sin/(sin/cos) = cos.

So, the final answer is cos(x).

Explanation

Step 1: sin/tan = sin/(sin/cos) = cos.

So, the final answer is cos(x).

13) Simplify: tan²(x)/tan(x)

Tan(x)

Tan³(x)

Cot(x)

Step 1: tan²/tan = tan (where defined).

So, the final answer is tan(x).

Explanation

Step 1: tan²/tan = tan (where defined).

So, the final answer is tan(x).

14) Simplify: sin(x)·cot(x) + cos(x)

Sin(x) + cos(x)

2cos(x)

Tan(x)

Sec(x)

Step 1: sin·cot = sin·(cos/sin) = cos.

Step 2: cos + cos = 2cos.

So, the final answer is 2cos(x).

Explanation

Step 1: sin·cot = sin·(cos/sin) = cos.

Step 2: cos + cos = 2cos.

So, the final answer is 2cos(x).

15) Which identity is correct for tan(θ)?

Tan(θ) = cos(θ)/sin(θ)

Tan(θ) = sin(θ)/sec(θ)

Tan(θ) = sin(θ)·cos(θ)

Tan(θ) = 1/cot(θ) only

Step 1: By definition, tanθ = sinθ/cosθ.

So, the final answer is B.

Explanation

Step 1: By definition, tanθ = sinθ/cosθ.

So, the final answer is B.

16) Which expression is NOT equivalent to sin²(θ)/cos(θ)?

Sin(θ)·tan(θ)

Tan(θ)·sin(θ)

Sin(θ)/cot(θ)

Sin³(θ)/cos²(θ)

Step 1: sinθ·tanθ = sinθ·(sinθ/cosθ) = sin²θ/cosθ (equivalent).

Step 2: sinθ/cotθ = sinθ/(cosθ/sinθ) = sin²θ/cosθ (equivalent).

Step 3: sin³θ/cos²θ ≠ sin²θ/cosθ (extra sinθ/cosθ factor).

So, the final answer is D.

Explanation

17) Simplify: tan(x)/sin(x)

Sec(x)

Csc(x)

Cos(x)

Sin(x)

Step 1: tanx/sinx = (sinx/cosx)/sinx = 1/cosx.

Step 2: 1/cosx = secx.

So, the final answer is secx.

Explanation

Step 1: tanx/sinx = (sinx/cosx)/sinx = 1/cosx.

Step 2: 1/cosx = secx.

So, the final answer is secx.

18) If cot(β) = 4/3, what is tan(β)?

4/3

3/4

7/3

5/3

Step 1: tanβ = 1/cotβ = 1/(4/3) = 3/4.

So, the final answer is 3/4.

Explanation

Step 1: tanβ = 1/cotβ = 1/(4/3) = 3/4.

So, the final answer is 3/4.

19) Simplify: (1 + tan²(x))/tan(x)

Csc(x)·sec(x)

Tan(x)

Cot(x)

Sin(x)

Step 1: 1 + tan²x = sec²x.

Step 2: sec²x / tanx = (1/cos²x)/(sin(x)/cos(x) = 1/(sin(x) cos(x) = csc(x)·sec(x).

So, the final answer is csc(x)·sec(x).

Explanation

Step 1: 1 + tan²x = sec²x.

Step 2: sec²x / tanx = (1/cos²x)/(sin(x)/cos(x) = 1/(sin(x) cos(x) = csc(x)·sec(x).

So, the final answer is csc(x)·sec(x).

20) If sin(θ)/cos(θ) = k, what is cos(θ)/sin(θ) in terms of k?

−k

K²

1/k

Step 1: sinθ/cosθ = k ⇒ cosθ/sinθ = 1/k (reciprocal).

So, the final answer is 1/k.

Explanation

Step 1: sinθ/cosθ = k ⇒ cosθ/sinθ = 1/k (reciprocal).

So, the final answer is 1/k.

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Cierra Henderson |MBA |

K-12 Expert

Cierra is an educational consultant and curriculum developer who has worked with students in K-12 for a variety of subjects including English and Math as well as test prep. She specializes in one-on-one support for students especially those with learning differences. She holds an MBA from the University of Massachusetts Amherst and a certificate in educational consulting from UC Irvine.