In trigonometry, if cos θ is given by the expression (x^2 − y^2) / (x^2 + y^2) for real numbers x and y, then what is the value of cot θ expressed purely in terms of x and y?

Introduction / Context:
This question tests a standard trigonometric parametrisation where cosine is written in terms of two real parameters x and y. The form cos θ = (x^2 − y^2) / (x^2 + y^2) is closely related to the tangent half angle formulas and to expressing sine and cosine using rational functions. The goal is to convert this information into cot θ, which is the ratio of cosine to sine, written only in terms of x and y.

Given Data / Assumptions:

• cos θ = (x^2 − y^2) / (x^2 + y^2).
• θ is an acute angle so sine and cosine are positive.
• x and y are real numbers, and x^2 + y^2 ≠ 0.
• We need cot θ, which is cos θ / sin θ, expressed using x and y only.

Concept / Approach:
The key idea is to first compute sin θ from the identity sin^2 θ + cos^2 θ = 1. Once we obtain sin θ in terms of x and y, we can form cot θ = cos θ / sin θ. Because cos θ already has denominator x^2 + y^2, we will square it, subtract from 1, and then take the square root to get sin θ. Simplification will produce a clean expression involving the product xy and the difference x^2 − y^2.

Step-by-Step Solution:
1. Start from cos θ = (x^2 − y^2) / (x^2 + y^2). 2. Compute cos^2 θ = (x^2 − y^2)^2 / (x^2 + y^2)^2. 3. Use sin^2 θ = 1 − cos^2 θ. 4. Substitute: sin^2 θ = 1 − (x^2 − y^2)^2 / (x^2 + y^2)^2. 5. Write 1 as (x^2 + y^2)^2 / (x^2 + y^2)^2. 6. Then sin^2 θ = [(x^2 + y^2)^2 − (x^2 − y^2)^2] / (x^2 + y^2)^2. 7. Expand numerators: (x^2 + y^2)^2 = x^4 + 2x^2 y^2 + y^4, (x^2 − y^2)^2 = x^4 − 2x^2 y^2 + y^4. 8. Subtract: (x^4 + 2x^2 y^2 + y^4) − (x^4 − 2x^2 y^2 + y^4) = 4x^2 y^2. 9. So sin^2 θ = 4x^2 y^2 / (x^2 + y^2)^2. 10. For an acute θ and real x, y, take the positive root: sin θ = 2xy / (x^2 + y^2). 11. Now form cot θ = cos θ / sin θ. 12. Substitute cos θ and sin θ: cot θ = [(x^2 − y^2) / (x^2 + y^2)] / [2xy / (x^2 + y^2)]. 13. Cancel x^2 + y^2 in numerator and denominator. 14. This gives cot θ = (x^2 − y^2) / (2xy).

Verification / Alternative check:
We can cross check using the tangent half angle substitution t = tan(θ / 2). In that parametrisation, cos θ = (1 − t^2) / (1 + t^2) and sin θ = 2t / (1 + t^2). If we identify t = y / x, then cos θ becomes (x^2 − y^2) / (x^2 + y^2) and sin θ becomes 2xy / (x^2 + y^2). This matches our derived expressions and confirms that cot θ = (x^2 − y^2) / (2xy).

Why Other Options Are Wrong:
Option 2: 2xy / (x^2 − y^2) is equal to tan θ, not cot θ, because it is sin θ / cos θ.
Option 3: (x^2 + y^2) / (2xy) is the reciprocal of sin θ, which would correspond to cosec θ, not cot θ.
Option 4: (x^2 + y^2) / (x^2 − y^2) is the reciprocal of cos θ, that is sec θ, not cot θ.

Common Pitfalls:
A common mistake is to forget to square the denominator when squaring cos θ, which leads to an incorrect expression for sin^2 θ. Another frequent error is mixing up tan θ and cot θ when forming cos θ / sin θ versus sin θ / cos θ. Students may also incorrectly cancel terms before properly simplifying the fraction, especially with (x^2 + y^2) terms. Careful algebra at each step avoids these errors.

Final Answer:
Therefore, the correct value of cot θ in terms of x and y is (x^2 − y^2) / (2xy).