Proving the Pythagorean Identity Quiz

11th Grade

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| Attempts: 41 | Questions: 20 | Updated: Jan 22, 2026

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1) Which expression equals 1 by the Pythagorean identity?

Sin²θ + cos²θ

Tan²θ + 1

Sin²θ − cos²θ

Sec²θ − 1

The Pythagorean identity states sin²θ + cos²θ = 1.

So final answer is sin²θ + cos²θ.

Explanation

The Pythagorean identity states sin²θ + cos²θ = 1.

So final answer is sin²θ + cos²θ.

Submit

About This Quiz

Proving The Pythagorean Identity Quiz - Quiz

Go from “know it” to “show it.” You’ll prove sin²θ + cos²θ = 1 from the unit circle, generate the derived forms (1 + tan²θ = sec²θ and 1 + cot²θ = csc²θ), and apply them to compute exact trig values. Clear steps, justified algebra, and solid geometry backing.

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2) If sinθ = 15/17 and θ is in Quadrant I, what is tanθ?

8/15

−8/15

−15/8

15/8

Given: sinθ = 15/17, θ in QI.

Step 1: cos²θ = 1 − (15/17)² = 64/289 cosθ = 8/17.

Step 2: tanθ = sinθ / cosθ = (15/17)/(8/17) = 15/8.

So final answer is 15/8.

Explanation

Given: sinθ = 15/17, θ in QI.

Step 1: cos²θ = 1 − (15/17)² = 64/289 cosθ = 8/17.

Step 2: tanθ = sinθ / cosθ = (15/17)/(8/17) = 15/8.

So final answer is 15/8.

Submit

3) If sinθ = 5/13 and θ is in Quadrant II, which is true?

Cosθ = −12/13 and tanθ = −5/12

Cosθ = 12/13 and tanθ = 5/12

Cosθ = −12/13 and tanθ = 5/12

Cosθ = 12/13 and tanθ = −5/12

Given: sinθ = 5/13, θ in QII.

Step 1: cos²θ = 1 − (25/169) = 144/169 cosθ = −12/13 (QII).

Step 2: tanθ = sinθ / cosθ = (5/13)/(−12/13) = −5/12.

So final answer is A. cosθ = −12/13 and tanθ = −5/12.

Explanation

Submit

4) Which reasoning best completes a proof: On the unit circle, a point corresponding to angle θ is (cosθ, sinθ). By the equation of the unit circle, x² + y² = 1, it follows that:

Sin²θ + cos²θ = 1

Sin²θ − cos²θ = 1

Tan²θ + 1 = 1

Sinθ + cosθ = 1

Substitute x = cosθ and y = sinθ into x² + y² = 1 sin²θ + cos²θ = 1.

So final answer is sin²θ + cos²θ = 1.

Explanation

Substitute x = cosθ and y = sinθ into x² + y² = 1 sin²θ + cos²θ = 1.

So final answer is sin²θ + cos²θ = 1.

Submit

5) Which of the following is a correct rearrangement of the Pythagorean identity?

Sin²θ = 1 + cos²θ

1 = sinθ + cosθ

Sin²θ − cos²θ = 1

Cos²θ = 1 − sin²θ

From sin²θ + cos²θ = 1 rearranging gives cos²θ = 1 − sin²θ.

So final answer is cos²θ = 1 − sin²θ.

Explanation

From sin²θ + cos²θ = 1 rearranging gives cos²θ = 1 − sin²θ.

So final answer is cos²θ = 1 − sin²θ.

Submit

6) If secθ = −5/4 and θ is in Quadrant II, find sinθ.

3/5

−3/5

−4/5

4/5

Given: secθ = −5/4 cosθ = −4/5.

Step 1: sin²θ = 1 − (−4/5)² = 1 − 16/25 = 9/25 sinθ = ±3/5.

Step 2: QII ⇒ sine positive sinθ = 3/5.

So final answer is 3/5.

Explanation

Given: secθ = −5/4 cosθ = −4/5.

Step 1: sin²θ = 1 − (−4/5)² = 1 − 16/25 = 9/25 sinθ = ±3/5.

Step 2: QII ⇒ sine positive sinθ = 3/5.

So final answer is 3/5.

Submit

7) Which step helps prove sin²θ + cos²θ = 1 from the unit circle definition?

Using tanθ = sinθ/cosθ directly

Using the double-angle formula

Using x² + y² = r² with r = 1

Using x = r cosθ, y = r sinθ with r = 2

The unit circle is defined by x² + y² = r², and for r = 1, substituting x = cosθ, y = sinθ gives sin²θ + cos²θ = 1.

So final answer is x² + y² = r² with r = 1.

Explanation

The unit circle is defined by x² + y² = r², and for r = 1, substituting x = cosθ, y = sinθ gives sin²θ + cos²θ = 1.

So final answer is x² + y² = r² with r = 1.

Submit

8) Given cosθ = −5/13 and θ in Quadrant II, find tanθ.

12/5

−5/12

−12/5

5/12

Given: cosθ = −5/13, θ in QII.

Step 1: sin²θ = 1 − (25/169) = 144/169 sinθ = 12/13.

Step 2: tanθ = (12/13)/(−5/13) = −12/5.

So final answer is −12/5.

Explanation

Given: cosθ = −5/13, θ in QII.

Step 1: sin²θ = 1 − (25/169) = 144/169 sinθ = 12/13.

Step 2: tanθ = (12/13)/(−5/13) = −12/5.

So final answer is −12/5.

Submit

9) If cosθ = −12/13 and θ is in Quadrant III, what is sinθ?

5/13

12/13

−5/13

−12/13

Given: cosθ = −12/13, θ in QIII. Goal: Find sinθ.

Step 1: sin²θ = 1 − cos²θ = 1 − (144/169) = 25/169.

Step 2: sinθ = ±5/13.

Step 3: QIII ⇒ sine negative sinθ = −5/13.

So final answer is −5/13.

Explanation

Submit

10) If sinθ = −4/5 and θ is in Quadrant III, find tanθ.

−3/4

3/4

4/3

−4/3

Given: sinθ = −4/5, θ in QIII.

Step 1: cos²θ = 1 − (−4/5)² = 1 − 16/25 = 9/25 cosθ = −3/5.

Step 2: tanθ = sinθ / cosθ = (−4/5)/(−3/5) = 4/3.

So final answer is 4/3.

Explanation

Given: sinθ = −4/5, θ in QIII.

Step 1: cos²θ = 1 − (−4/5)² = 1 − 16/25 = 9/25 cosθ = −3/5.

Step 2: tanθ = sinθ / cosθ = (−4/5)/(−3/5) = 4/3.

So final answer is 4/3.

Submit

11) If tanθ = 7/24 and θ is in Quadrant I, what is cosθ?

7/25

24/25

−24/25

−7/25

Given: tanθ = 7/24, θ in QI.

Step 1: Reference triangle opp = 7, adj = 24, hyp = 25.

Step 2: cosθ = 24/25.

So final answer is 24/25.

Explanation

Given: tanθ = 7/24, θ in QI.

Step 1: Reference triangle opp = 7, adj = 24, hyp = 25.

Step 2: cosθ = 24/25.

So final answer is 24/25.

Submit

12) If cosθ = √3/2 and θ is in Quadrant IV, find sinθ.

√3/2

1/2

−1/2

−√3/2

Given: cosθ = √3/2, θ in QIV.

Step 1: sin²θ = 1 − (3/4) = 1/4 sinθ = ±1/2.

Step 2: QIV ⇒ sine negative sinθ = −1/2.

So final answer is −1/2.

Explanation

Given: cosθ = √3/2, θ in QIV.

Step 1: sin²θ = 1 − (3/4) = 1/4 sinθ = ±1/2.

Step 2: QIV ⇒ sine negative sinθ = −1/2.

So final answer is −1/2.

Submit

13) If sinθ = −√2/2 and θ is in Quadrant III, find cosθ and tanθ.

Cosθ = √2/2, tanθ = −1

Cosθ = √2/2, tanθ = 1

Cosθ = −√2/2, tanθ = 1

Cosθ = −√2/2, tanθ = −1

Given: sinθ = −√2/2, θ in QIII.

Step 1: cos²θ = 1 − (1/2) = 1/2 cosθ = ±√2/2.

Step 2: QIII ⇒ cosine negative cosθ = −√2/2.

Step 3: tanθ = sinθ / cosθ = (−√2/2)/(−√2/2) = 1.

So final answer is cosθ = −√2/2, tanθ = 1.

Explanation

Submit

14) Suppose cosθ = 2/3 and θ is in Quadrant I. Find tanθ.

√5/3

2/√5

√5/2

√5/4

Given: cosθ = 2/3, θ in QI. Goal: Find tanθ.

Step 1: sin²θ = 1 − (2/3)² = 5/9 sinθ = √5/3.

Step 2: tanθ = sinθ / cosθ = (√5/3)/(2/3) = √5/2.

So final answer is √5/2.

Explanation

Given: cosθ = 2/3, θ in QI. Goal: Find tanθ.

Step 1: sin²θ = 1 − (2/3)² = 5/9 sinθ = √5/3.

Step 2: tanθ = sinθ / cosθ = (√5/3)/(2/3) = √5/2.

So final answer is √5/2.

Submit

15) If tanθ = −3/4 and θ is in Quadrant II, find sinθ.

3/5

−3/5

−4/5

4/5

Given: tanθ = −3/4, θ in QII.

Reference triangle opp = 3, adj = −4, hyp = 5.

In QII, sine is positive sinθ = 3/5.

So final answer is 3/5.

Explanation

Given: tanθ = −3/4, θ in QII.

Reference triangle opp = 3, adj = −4, hyp = 5.

In QII, sine is positive sinθ = 3/5.

So final answer is 3/5.

Submit

16) From the identity sin²θ + cos²θ = 1, dividing by sin²θ (sinθ ≠ 0) yields:

1 + tan²θ = sec²θ

1 − cot²θ = csc²θ

Tan²θ − 1 = sec²θ

1 + cot²θ = csc²θ

Dividing both sides by sin²θ 1 + cot²θ = csc²θ.

So final answer is 1 + cot²θ = csc²θ.

Explanation

Dividing both sides by sin²θ 1 + cot²θ = csc²θ.

So final answer is 1 + cot²θ = csc²θ.

Submit

17) If sinθ = 3/5 and θ is in Quadrant II, what is cosθ?

4/5

−3/5

3/5

−4/5

Given: sinθ = 3/5, θ in QII. Goal: Find cosθ.

Step 1: cos²θ = 1 − (3/5)² = 1 − 9/25 = 16/25.

Step 2: cosθ = ±4/5.

Step 3: In QII, cosine is negative cosθ = −4/5.

So final answer is −4/5.

Explanation

Submit

18) If sinθ = −7/25 and θ is in Quadrant IV, find cosθ.

24/25

−24/25

7/25

−7/25

Given: sinθ = −7/25, θ in QIV. Goal: Find cosθ.

Step 1: cos²θ = 1 − (−7/25)² = 1 − 49/625 = 576/625.

Step 2: cosθ = ±24/25.

Step 3: QIV ⇒ cosine positive cosθ = 24/25.

So final answer is 24/25.

Explanation

Submit

19) Which equivalent identity follows from dividing sin²θ + cos²θ = 1 by cos²θ (assuming cosθ ≠ 0)?

Tan²θ − 1 = sec²θ

Tan²θ + 1 = sec²θ

1 − tan²θ = sec²θ

1 + tan²θ = csc²θ

Dividing both sides by cos²θ gives tan²θ + 1 = sec²θ.

So final answer is tan²θ + 1 = sec²θ.

Explanation

Dividing both sides by cos²θ gives tan²θ + 1 = sec²θ.

So final answer is tan²θ + 1 = sec²θ.

Submit

20) Using sin²θ + cos²θ = 1, which identity is also true?

1 + cot²θ = csc²θ

1 − tan²θ = sec²θ

1 + tan²θ = csc²θ

1 + tan²θ = sec²θ

Dividing both sides by cos²θ gives tan²θ + 1 = sec²θ.

So final answer is 1 + tan²θ = sec²θ.

Explanation

Dividing both sides by cos²θ gives tan²θ + 1 = sec²θ.

So final answer is 1 + tan²θ = sec²θ.

Submit

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All (20)
Unanswered ()
Answered ()

Which expression equals 1 by the Pythagorean identity?

If sinθ = 15/17 and θ is in Quadrant I, what is...

If sinθ = 5/13 and θ is in Quadrant II, which is true?

Which reasoning best completes a proof: On the unit circle, a point...

Which of the following is a correct rearrangement of the Pythagorean...

If secθ = −5/4 and θ is in Quadrant II, find...

Which step helps prove sin²θ + cos²θ = 1 from...

Given cosθ = −5/13 and θ in Quadrant II, find...

If cosθ = −12/13 and θ is in Quadrant III, what is...

If sinθ = −4/5 and θ is in Quadrant III, find...

If tanθ = 7/24 and θ is in Quadrant I, what is cosθ?

If cosθ = √3/2 and θ is in Quadrant IV, find...

If sinθ = −√2/2 and θ is in Quadrant III, find...

Suppose cosθ = 2/3 and θ is in Quadrant I. Find...

If tanθ = −3/4 and θ is in Quadrant II, find...

From the identity sin²θ + cos²θ = 1, dividing by...

If sinθ = 3/5 and θ is in Quadrant II, what is cosθ?

If sinθ = −7/25 and θ is in Quadrant IV, find...

Which equivalent identity follows from dividing sin²θ +...

Using sin²θ + cos²θ = 1, which identity is also...

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