(sqrt(3)-(1)/(sqrt(3)))^(2)

Find the direction cosines of the line which is perpendicular to the lines with direction ratios 4, 1, 3 and 2, -3, 1 . a) (1)/(sqrt(3)),(1)/(5sqrt(3)),(-7)/(5sqrt(3)) b) (5)/(sqrt(3)),(1)/(sqrt(3)),(7)/(5) c) (2)/(sqrt(3)),(5)/(2sqrt(3)),(1)/(7sqrt(3)) d) (1)/(sqrt(3)),(2)/(sqrt(3)),(-1)/(sqrt(3))

(1)/(sqrt(2)+sqrt(3))-(sqrt(3)+1)/(2+sqrt(3))+(sqrt(2)+1)/(2+2sqrt(2))

A triangle is inscribed in a circle.The vertices of the triangle divide the circle into three arcs of length 3,4 and 5 ,then the area of the triangle is equal to 1.(9sqrt(3)(1+sqrt(3)))/(pi^(2)) 2.(9sqrt(3)(sqrt(3)-1))/(pi^(2)) 3.(9sqrt(3)(1+sqrt(3)))/(2 pi^(2)) 4.(9sqrt(3)(sqrt(3)-1))/(2 pi^(2))

(sqrt(3)-1)+(1)/(2)(sqrt(3)-1)^(2)+(1)/(3)(sqrt(3)-1)^(3)+….oo

(sqrt(3)-1)+(1)/(2)(sqrt(3)-1)^(2)+(1)/(3)(sqrt(3)-1)^(3)+….oo

if x=sqrt(3)+(1)/(sqrt(3)) and y=sqrt(3)-(1)/(sqrt(3)) then x^(2)-y^(2) is

A square of side a lies above the X- axis and has one vertex at the origin . The side passing through the origin makes an angle pi//6 with the positive direction of X-axis .The equation of its diagonal not passing through the origin is y(sqrt(3)-1)-x(1-sqrt(3))=2a y(sqrt(3)+1) +x(1-sqrt(3))=2a y(sqrt(3)+1)+x(1+sqrt(3)) =2a y(sqrt(3)+1)+x(sqrt(3)-1)=2a

Find the value of (1)/(2+sqrt(3))+(1)/(2-sqrt(3))