The area of the circle and the area of a regular polygon of `n` sides and of perimeter equal to that of the circle are in the ratio of `tan(pi/n):pi/n` (b) `cos(pi/n):pi/n` `sinpi/n :pi/n` (d) `cot(pi/n):pi/n`

The area of the circle and the area of a regular polygon of n sides and of perimeter equal to that of the circle are in the ratio of tan((pi)/(n)):(pi)/(n)(b)cos((pi)/(n)):(pi)/(n)sin(pi)/(n):(pi)/(n)(d)cot((pi)/(n)):(pi)/(n)

If the number of sides of two regular polygons having the same prerimeter be n and 2n respectiely, prove that their areas are in the ratio 2cos pi/n: (1+cos pi/n)

the sum of the radii of inscribed and circumscribed circle of an n sides regular polygon of side a is (A) a/2 cot (pi/(2n)) (B) acot(pi/(2n)) (C) a/4 cos, pi/(2n)) (D) a cot (pi/n)

The area of a regular polygon of n sides is (where r is inradius,R is circumradius,and a is side of the triangle) (nR^(2))/(2)sin((2 pi)/(n))(b)nr^(2)tan((pi)/(n))(na^(2))/(4)cot(pi)/(n)(d)nR^(2)tan((pi)/(n))

The value of tan(n pi+(pi)/(3)),n in I is

The value of lim_( n to oo) (1)/(n) ( sin (pi/ n) + sin (2pi/n) + …+ sin (n pi/ n) ) is

lim_(n rarr oo)n sin\ (2 pi)/(3n)*cos\ (2 pi)/(3n)

The period of f(x)=sin((pi x)/(n-1))+cos((pi x)/(n)),n in Z,n>2

If inside a big circle exactly n(nlt=3) small circles, each of radius r , can be drawn in such a way that each small circle touches the big circle and also touches both its adjacent small circles, then the radius of big circle is r(1+cos e cpi/n) (b) ((1+tanpi/n)/(cospi/pi)) r[1+cos e c(2pi)/n] (d) (r[s inpi/(2n)+cos(2pi)/n]^2)/(sinpi/n)