Trigonometry & Inverse Trigonometry
Trigonometric identities
Grade 11
Question:
<p>If \(\sec A \tan B + \tan A \sec B = 91\), then the value of \((\sec A \sec B + \tan A \tan B)^2\) is equal to:</p>
Step-by-Step Solution
Key Concept: Use the algebraic identity \(a^2 - b^2 = (a+b)(a-b)\) combined with the trigonometric identity \(\sec^2 \theta - \tan^2 \theta = 1\).
<p><strong>Step 1:</strong> Let \(x = \sec A \sec B + \tan A \tan B\) and \(y = \sec A \tan B + \tan A \sec B = 91\).</p><p><strong>Step 2:</strong> We observe that \(x^2 - y^2 = (\sec A \sec B + \tan A \tan B)^2 - (\sec A \tan B + \tan A \sec B)^2\).</p><p><strong>Step 3:</strong> Expanding: \(x^2 - y^2 = \sec^2 A \sec^2 B + \tan^2 A \tan^2 B + 2\sec A \sec B \tan A \tan B - \sec^2 A \tan^2 B - \tan^2 A \sec^2 B - 2\sec A \tan B \tan A \sec B\).</p><p><strong>Step 4:</strong> Simplifying: \(x^2 - y^2 = \sec^2 A (\sec^2 B - \tan^2 B) - \tan^2 A (\sec^2 B - \tan^2 B) = (\sec^2 A - \tan^2 A)(\sec^2 B - \tan^2 B) = 1 \cdot 1 = 1\).</p><p><strong>Step 5:</strong> Therefore, \(x^2 = y^2 + 1 = 91^2 + 1 = 8281 + 1 = 8282\).</p><p>∴ Answer is <strong>8282</strong>.</p>
Correct Answer: 8282