answer:Starting with the expression: [cot frac{x}{3} - cot 3x = frac{cos frac{x}{3}}{sin frac{x}{3}} - frac{cos 3x}{sin 3x}.] Combine the terms over a common denominator: [frac{cos frac{x}{3} sin 3x - sin frac{x}{3} cos 3x}{sin frac{x}{3} sin 3x}.] Apply the angle subtraction identity sin(a - b) = sin a cos b - cos a sin b: [frac{sin (3x - frac{x}{3})}{sin frac{x}{3} sin 3x} = frac{sin frac{8x}{3}}{sin frac{x}{3} sin 3x}.] Thus, k = boxed{frac{8}{3}}.

answer:Consider the tetrahedron (A_{1} A_{2} A_{3} A_{4}) where each pair of vertices is a center of a sphere that is tangent to the other three spheres. Given that there exists a point (O) such that a sphere with center (O) and radius (r) tangentially intersects with the four spheres centered at (A_{1}, A_{2}, A_{3}, A_{4}), and another sphere with center (O) and radius (R) intersects the six edges of the tetrahedron. 1. Suppose (O) is the center of a sphere with radius (R) that intersects the edges (A_{2}A_{3}, A_{3}A_{1},) and (A_{2}A_{1}) at points (P_{1}, P_{2},) and (P_{3}) respectively. 2. Let (r_1, r_2, r_3,) and (r_4) be the radii of the spheres centered at (A_{1}, A_{2}, A_{3},) and (A_{4}) respectively. 3. It is known that the distances from the vertices to the points of intersection (A_{3}P_{1}=A_{3}P_{2}=r_{3}, A_{4}P_{3}=A_{4}P_{2}=r_{4},) and (A_{2}P_{1}=A_{2}P_{3}=r_{2}). 4. Therefore, (P_{1}, P_{2}, P_{3}) are points of tangency of the spheres centered at (A_{2}, A_{3},) and (A_{4}). Let (Q_3) be the point of tangency of the sphere centered at (O) with the sphere centered at (A_{3}). **Case 1: Sphere (O) and sphere (A_{3}) are externally tangent** 5. Using the power of a point theorem (Intersecting Chords Theorem), we have: [ R^2 = OP_{1}^2 = OQ_{3} cdot (OA_{3} + r_{3}) = r(r + 2r_{3}) ] 6. Similarly for sphere (A_{2}): [ R^2 = r(r + 2r_{2}) ] 7. Equating both expressions: [ r(r + 2r_{3}) = r(r + 2r_{2}) ] Therefore, (r_{2} = r_{3}). 8. By symmetry and a similar reasoning, we also have (r_{4} = r_{3}) and (r_{1} = r_{3}). 9. Therefore, (r_{1} = r_{2} = r_{3} = r_{4}). Thus, all the radii (r_{1}, r_{2}, r_{3},) and (r_{4}) are equal, implying that all edges of the tetrahedron are equal. 10. Hence, the tetrahedron (A_{1} A_{2} A_{3} A_{4}) is a regular tetrahedron. **Case 2: Sphere (O) and sphere (A_{3}) are internally tangent** 11. In this case, again using the power of a point theorem, we have: [ R^2 = OP_{1}^2 = OQ_{3} cdot (OA_{3} - r_{3}) = r(r - 2r_{3}) ] The subsequent steps and logic will mirror Case 1, and we conclude: 12. Hence, in any case, the tetrahedron (A_{1} A_{2} A_{3} A_{4}) is a regular tetrahedron. Conclusion: [ boxed{text{The tetrahedron } A_{1} A_{2} A_{3} A_{4} text{ is regular.}} ]

answer:For the line l_1: ax+2y+6=0, and the line l_2: x+(a-1)y+a^2-1=0, since l_1 perp l_2, we have a cdot 1 + 2(a-1) = 0; Solving this, we get: a= boxed{frac {2}{3}}. Therefore, the correct choice is: B. Based on the perpendicularity of lines l_1 and l_2, by using the formula A_1 cdot A_2 + B_1 cdot B_2 = 0, we can set up an equation to find the value of a. This problem tests the application of the equation of a line and the concept of two lines being perpendicular to each other, which is a fundamental question.

answer:Since 4x^2-5xy+4y^2=5, we have 5xy=4x^2+4y^2-5. Also, since 2xy leq x^2+y^2, we get 5xy=4x^2+4y^2-5 leq frac{5}{2}(x^2+y^2). Let S=x^2+y^2, then 4S-5 leq frac{5}{2}S. Therefore, S leq frac{10}{3}, which means boxed{S_{max} = frac{10}{3}}. Since x^2+y^2 geq -2xy, we have 5xy=4x^2+4y^2-5 geq -8xy-5. Thus, xy leq -frac{5}{13}, and hence -xy geq frac{5}{13}. Therefore, S=x^2+y^2 geq -2xy geq frac{10}{13}, which means boxed{S_{min} = frac{10}{13}}. Thus, frac{1}{S_{max}} + frac{1}{S_{min}} = frac{3}{10} + frac{13}{10} = boxed{frac{8}{5}}. The answer is: boxed{frac{8}{5}}. By utilizing the inequality 2xy leq x^2+y^2, we can derive that 5xy=4x^2+4y^2-5 leq frac{5}{2}(x^2+y^2), which allows us to find the maximum value of s. Similarly, from x^2+y^2 geq -2xy and 5xy=4x^2+4y^2-5 geq -8xy-5, we can determine the range of xy, and thus find the minimum value of s. Substituting these values gives us the solution. This problem primarily tests the application of basic inequalities in solving for extremal values, with the key being the flexible use of basic formulas.