Trigonometry & Inverse Trigonometry
Trigonometric identities
Grade 11
Question:
<p>\(\sec^2 \theta = \frac{4xy}{(x+y)^2}\) is true if and only if</p>
<p>(a) \(x + y \neq 0\)</p>
<p>(b) \(x = y\), \(x \neq 0\)</p>
<p>(c) \(x = y\)</p>
<p>(d) \(x \neq 0\), \(y \neq 0\)</p>
Step-by-Step Solution
Key Concept: For sec²θ to equal a specific value, that value must be ≥ 1 (since sec²θ ≥ 1 always). Apply AM-GM inequality to find when (x+y)²/(4xy) ≥ 1, which determines the constraint on x and y.
<p><strong>Step 1:</strong> Recognize that for any real angle θ: sec²θ ≥ 1 (since |secθ| ≥ 1)</p><p><strong>Step 2:</strong> Therefore: $\frac{4xy}{(x+y)^2} \geq 1$ is required</p><p><strong>Step 3:</strong> This gives: $4xy \geq (x+y)^2$</p><p><strong>Step 4:</strong> Expanding: $4xy \geq x^2 + 2xy + y^2$</p><p><strong>Step 5:</strong> Simplifying: $0 \geq x^2 - 2xy + y^2 = (x-y)^2$</p><p><strong>Step 6:</strong> Since $(x-y)^2 \geq 0$ always, equality holds only when $(x-y)^2 = 0$</p><p><strong>Step 7:</strong> Therefore: $x = y$ (and both must be positive for sec²θ to be positive)</p><p>∴ Answer: <strong>x = y and x, y > 0</strong> (or equivalent condition)</p>
Correct Answer: B