<p>We have \[\Delta_1 = \begin{vmatrix} x & \sin\theta & \cos\theta \\ -\sin\theta & -x & 1 \\ \cos\theta & 1 & x \end{vmatrix}\] and \[\Delta_2 = \begin{vmatrix} x & \sin 2\theta & \cos 2\theta \\ -\sin 2\theta & -x & 1 \\ \cos 2\theta & 1 & x \end{vmatrix}\] Then \(\Delta_1 + \Delta_2\) equals:</p>
Step-by-Step Solution
Key Concept: Recognize the symmetric structure in both determinants and use row/column operations to reveal that each determinant can be expressed as a sum of simpler terms, or compute directly using the pattern that emerges from expanding both determinants.
<p><strong>Step 1:</strong> Expand Δ₁ along the first row:</p><p>Δ₁ = x|−x 1| − sin θ|−sin θ 1| + cos θ|−sin θ −x|</p><p> |1 x| |cos θ x| |cos θ 1|</p><p>= x(−x² − 1) − sin θ(−x sin θ − cos θ) + cos θ(−sin θ − (−x cos θ))</p><p>= −x³ − x + x sin² θ + sin θ cos θ − sin θ cos θ + x cos² θ</p><p>= −x³ − x + x(sin² θ + cos² θ)</p><p>= −x³ − x + x = <strong>−x³</strong></p><p><strong>Step 2:</strong> By the same process for Δ₂ (replacing θ with 2θ):</p><p>Δ₂ = −x³ − x + x(sin² 2θ + cos² 2θ) = −x³ − x + x = <strong>−x³</strong></p><p><strong>Step 3:</strong> Add the results:</p><p>Δ₁ + Δ₂ = −x³ + (−x³) = <strong>−2x³</strong></p><p>∴ Answer: <strong>A</strong></p>
Correct Answer: A