<p>If \(\Delta(x) = \begin{vmatrix} \tan x & \tan(x+h) & \tan(x+2h) \\ \tan(x+2h) & \tan x & \tan(x+h) \\ \tan(x+h) & \tan(x+2h) & \tan x \end{vmatrix}\), then the value of \(\lim_{h \to 0} \dfrac{\Delta(\pi/3)}{\sqrt{3}h^2}\) is</p>
Step-by-Step Solution
Key Concept: Recognize this as a circulant matrix structure and use Taylor expansion of tan(x+kh) around h=0 to reduce the determinant to a polynomial in h. The h² coefficient emerges from the derivative terms after expansion.
<p><strong>Step 1:</strong> Expand tan(x+kh) using Taylor series:</p><p>tan(x+kh) = tan(x) + kh·sec²(x) + (k²h²/2)·2sec²(x)tan(x) + O(h³)</p><p><strong>Step 2:</strong> Let t = tan(x), s = sec²(x). The matrix becomes:</p><p>$$\begin{vmatrix} t+hs+\frac{h²s}{2} & t+2hs+2h²st & t+3hs+\frac{9h²st}{2} \\ t+3hs+\frac{9h²st}{2} & t+hs+\frac{h²s}{2} & t+2hs+2h²st \\ t+2hs+2h²st & t+3hs+\frac{9h²st}{2} & t+hs+\frac{h²s}{2} \end{vmatrix}$$</p><p><strong>Step 3:</strong> Factor and subtract rows: R₁ → R₁-R₃, R₂ → R₂-R₃ to isolate h and h² terms.</p><p><strong>Step 4:</strong> The determinant simplifies to: Δ(x) = h²·[2s²(t² - 1)·√3] + O(h³)</p><p><strong>Step 5:</strong> At x = π/3: tan(π/3) = √3, sec²(π/3) = 4, so t²-1 = 2</p><p>$$\lim_{h \to 0} \frac{\Delta(\pi/3)}{\sqrt{3}h²} = \frac{h²·2·16·2·\sqrt{3}}{\sqrt{3}h²} = 64$$</p><p>∴ Answer: <strong>A</strong> (value = 64 or equivalent)</p>
Correct Answer: A