Reciprocal Identities Application Quiz

  • 11th Grade
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| Questions: 20 | Updated: Dec 11, 2025
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1) Simplify: sinθ/cscθ + cosθ/secθ

Explanation

Step 1: Rewrite the expression using reciprocal identities.

The given expression is (sinθ / cscθ) + (cosθ / secθ).

We know cscθ = 1/sinθ and secθ = 1/cosθ.

Substitute these identities: (sinθ / (1/sinθ)) + (cosθ / (1/cosθ)).

Step 2: Simplify each term.

Multiply numerator by reciprocal of the denominator: (sinθ × sinθ) + (cosθ × cosθ) = sin²θ + cos²θ.

Step 3: Apply the Pythagorean identity.

sin²θ + cos²θ = 1.

So, the final answer is 1.

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About This Quiz
Reciprocal Identities Application Quiz - Quiz

Turn definitions into shortcuts. Expect quick simplifications (like sin·csc = 1), sign checks by quadrant, and triangle/unit-circle setups to evaluate sec, csc, and cot. You’ll choose the fastest path from mixed forms to a single, clean trig result.

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2) Simplify the expression using reciprocal identities: sinθ · cscθ

Explanation

Step 1: Use the reciprocal identity for cscθ = 1/sinθ.

Step 2: Substitute into the expression: sinθ × (1/sinθ) = 1.

So, the final answer is 1.

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3) Use reciprocal identities to simplify: secθ · cosθ

Explanation

Step 1: secθ = 1/cosθ.

Step 2: (1/cosθ) × cosθ = 1.

So, the final answer is 1.

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4) Rewrite tanθ using only sine and cosine.

Explanation

Step 1: tanθ = sinθ/cosθ.

So, the final answer is sinθ/cosθ.

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5) Simplify: secθ·cosθ + cscθ·sinθ

Explanation

Step 1: secθ·cosθ = 1 and cscθ·sinθ = 1.

Step 2: 1 + 1 = 2.

So, the final answer is 2.

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6) If cscθ = 13/5 and θ in QII, what is sinθ?

Explanation

Step 1: sinθ = 1/cscθ = 5/13.

Step 2: Since θ is in Quadrant II, sinθ is positive.

So, the final answer is 5/13.

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7) Simplify: cscθ − 1/sinθ

Explanation

Step 1: cscθ = 1/sinθ.

Step 2: 1/sinθ − 1/sinθ = 0.

So, the final answer is 0.

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8) If tanθ = 7, what is cotθ?

Explanation

Step 1: cotθ = 1/tanθ = 1/7.

So, the final answer is 1/7.

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9) Which simplifies to 1 for all θ where defined?

Explanation

Step 1: Use reciprocal identity cotθ = 1/tanθ.

Step 2: Substitute: tanθ × (1/tanθ) = 1.

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10) If csc α = 5/13 and α is acute, find sec α.

Explanation

Step 1: sinα = 1/cscα = 13/5 → This is reversed. Actually, sinα = 1/(5/13) = 13/5 is incorrect. Let’s fix it.

Step 1: sinα = 1/cscα = 1 / (5/13) = 13/5 (impossible, greater than 1).

Hence cscα = 13/5 instead of 5/13.

Now sinα = 5/13.

Step 2: Use the Pythagorean identity cos²α + sin²α = 1.

cos²α = 1 − (5/13)² = 1 − 25/169 = 144/169 → cosα = 12/13.

Step 3: secα = 1/cosα = 13/12.

So, the final answer is 13/12.

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11) Right triangle: opposite = 3, hypotenuse = 6. Which equals cscθ?

Explanation

Step 1: sinθ = opposite / hypotenuse = 3/6 = 1/2.

Step 2: cscθ = 1/sinθ = 1 / (1/2) = 2.

So, the final answer is 2.

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12) Unit circle point P(−a, −b) in QIII (a,b>0). Which is true?

Explanation

Step 1: In Quadrant III, both sine and cosine are negative.

Step 2: secθ = 1/cosθ → negative; cscθ = 1/sinθ → negative.

So, the final answer is secθ < 0 and cscθ < 0.

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13) Simplify: 1 + tanθ · cotθ

Explanation

Step 1: Use cotθ = 1/tanθ.

Step 2: Substitute: 1 + tanθ × (1/tanθ) = 1 + 1 = 2.

So, the final answer is 2.

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14) Simplify: sinθ/(1/cscθ)

Explanation

Step 1: Replace 1/cscθ with sinθ.

Step 2: The expression becomes sinθ × sinθ = sin²θ.

So, the final answer is sin²θ.

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15) If cosθ = −3/5 and θ in QII, find secθ.

Explanation

Step 1: secθ = 1/cosθ = 1 / (−3/5) = −5/3.

Step 2: In Quadrant II, cos is negative, so sec is also negative.

So, the final answer is −5/3.

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16) If sinθ = −4/7 and θ in QIII, find cscθ.

Explanation

Step 1: Use reciprocal identity cscθ = 1/sinθ.

Step 2: Substitute: 1 / (−4/7) = −7/4.

So, the final answer is −7/4.

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17) Simplify: (1 + secθ)/secθ

Explanation

Step 1: Rewrite secθ as 1/cosθ.

Step 2: Substitute: (1 + 1/cosθ) / (1/cosθ).

Step 3: Multiply numerator and denominator by cosθ:

[(cosθ + 1)/cosθ] × cosθ = 1 + cosθ.

So, the final answer is (1 + cosθ)/cosθ.

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18) Solve tan x = √3, 0 ≤ x < 2π.

Explanation

Step 1: tanx = √3 means reference angle = π/3.

Step 2: Tangent is positive in Quadrants I and III.

So, x = π/3 and 4π/3.

So, the final answer is x = π/3 and 4π/3.

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19) Simplify (secθ + tanθ)(secθ − tanθ)

Explanation

Step 1: Use difference of squares formula.

(secθ + tanθ)(secθ − tanθ) = sec²θ − tan²θ.

Step 2: Use the identity sec²θ − tan²θ = 1.

So, the final answer is 1.

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20) A ladder leans against a wall making an angle θ with the ground. The angle satisfies tanθ = 3/4. Find cotθ and secθ.

Explanation

Step 1: tanθ = 3/4 ⇒ opposite = 3, adjacent = 4, hypotenuse = 5.

Step 2: cotθ = adjacent / opposite = 4/3 and secθ = hypotenuse / adjacent = 5/4.

So, the final answer is cotθ = 4/3 and secθ = 5/4.

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Simplify: sinθ/cscθ + cosθ/secθ
Simplify the expression using reciprocal identities: sinθ...
Use reciprocal identities to simplify: secθ · cosθ
Rewrite tanθ using only sine and cosine.
Simplify: secθ·cosθ + cscθ·sinθ
If cscθ = 13/5 and θ in QII, what is sinθ?
Simplify: cscθ − 1/sinθ
If tanθ = 7, what is cotθ?
Which simplifies to 1 for all θ where defined?
If csc α = 5/13 and α is acute, find sec α.
Right triangle: opposite = 3, hypotenuse = 6. Which equals cscθ?
Unit circle point P(−a, −b) in QIII (a,b>0). Which is...
Simplify: 1 + tanθ · cotθ
Simplify: sinθ/(1/cscθ)
If cosθ = −3/5 and θ in QII, find secθ.
If sinθ = −4/7 and θ in QIII, find cscθ.
Simplify: (1 + secθ)/secθ
Solve tan x = √3, 0 ≤ x < 2π.
Simplify (secθ + tanθ)(secθ − tanθ)
A ladder leans against a wall making an angle θ with the ground....
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