Question:
If
\(S = \left\{x \in [0, 2\pi] : \begin{vmatrix} 0 \& \cos x \& -\sin x \\ \sin x \& 0 \& \cos x \\ \cos x \& \sin x \& 0 \end{vmatrix} = 0\right\}\),
then \(\sum_{x \in S} \tan\left(\frac{\pi}{3} + x\right)\) is equal to :
If
\(S = \left\{x \in [0, 2\pi] : \begin{vmatrix} 0 \& \cos x \& -\sin x \\ \sin x \& 0 \& \cos x \\ \cos x \& \sin x \& 0 \end{vmatrix} = 0\right\}\),
then \(\sum_{x \in S} \tan\left(\frac{\pi}{3} + x\right)\) is equal to :
\(S = \left\{x \in [0, 2\pi] : \begin{vmatrix} 0 \& \cos x \& -\sin x \\ \sin x \& 0 \& \cos x \\ \cos x \& \sin x \& 0 \end{vmatrix} = 0\right\}\),
then \(\sum_{x \in S} \tan\left(\frac{\pi}{3} + x\right)\) is equal to :
Show Hint
The determinant evaluates to \(\cos^3 x - \sin^3 x\), so \(\tan x = 1\); sum \(\tan(60^\circ + x)\) over the two roots.
Updated On: Nov 11, 2026
- \(4 + 2\sqrt{3}\)
- \(-2 + \sqrt{3}\)
- \(-2 - \sqrt{3}\)
- \(-4 - 2\sqrt{3}\)
Show Solution
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The Correct Option is D
Solution and Explanation
Expand along the first row: \(0 - \cos x(0 - \cos^2 x) + (-\sin x)(\sin^2 x - 0) = \cos^3 x - \sin^3 x\).
Setting it to 0 gives \(\cos x = \sin x\), i.e. \(\tan x = 1\), so in \([0, 2\pi]\): \(x = \frac{\pi}{4}, \frac{5\pi}{4}\).
Both give the same value (tan has period \(\pi\)): \(\tan\left(\frac\pi3 + \frac\pi4\right) = \frac{\sqrt3 + 1}{1 - \sqrt3} = \frac{(\sqrt3+1)^2}{(1-\sqrt3)(1+\sqrt3)} = \frac{4 + 2\sqrt3}{-2} = -(2 + \sqrt3)\).
Sum \(= 2 \times (-(2+\sqrt3)) = -4 - 2\sqrt3\).
Answer: (4) \(-4 - 2\sqrt{3}\)
Setting it to 0 gives \(\cos x = \sin x\), i.e. \(\tan x = 1\), so in \([0, 2\pi]\): \(x = \frac{\pi}{4}, \frac{5\pi}{4}\).
Both give the same value (tan has period \(\pi\)): \(\tan\left(\frac\pi3 + \frac\pi4\right) = \frac{\sqrt3 + 1}{1 - \sqrt3} = \frac{(\sqrt3+1)^2}{(1-\sqrt3)(1+\sqrt3)} = \frac{4 + 2\sqrt3}{-2} = -(2 + \sqrt3)\).
Sum \(= 2 \times (-(2+\sqrt3)) = -4 - 2\sqrt3\).
Answer: (4) \(-4 - 2\sqrt{3}\)
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