The area of a regular polygon of `n` sides is (where `r` is inradius, `R` is circumradius, and `a` is side of the triangle) `(n R^2)/2sin((2pi)/n)` (b) `n r^2tan(pi/n)` `(n a^2)/4cotpi/n` (d) `n R^2tan(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 (a) tan(pi/n):pi/n (b) cos(pi/n):pi/n (c) sin(pi/n) :pi/n (d) cot(pi/n):pi/n

Prove that the area of a regular polygon hawing 2n sides, inscribed in a circle, is the geometric mean of the areas of the inscribed and circumscribed polygons of n sides.

A B C is an equilateral triangle of side 4c mdot If R ,ra n dh are the circumradius, inradius, and altitude, respectively, then (R+h)/h is equal to 4 (b) 2 (c) 1 (d) 3

The value of lim_(n->oo) [tan(pi/(2n)) tan((2pi)/(2n))........tan((npi)/(2n))]^(1/n) is

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[sin pi/(2n)+cospi/(2n)]^2)/(sinpi/n)

The sum of series Sigma_(r=0)^(r) (-1)^r(n+2r)^2 (where n is even) is

Prove that: int_0^(2pi)(xsin^(2n)x)/(sin^(2n)+cos^(2n)x)dx = pi^2

The value of (r(n+2))/(r(n))=90 then n is

The general solution of the equation sin^(100)x-cos^(100)x=1 is 2npi+pi/3,n in I (b) n pi+pi/2,n in I npi+pi/4,n in I (d) 2npi=pi/3,n in I

If tanx=ntany ,n in R^+, then the maximum value of sec^2(x-y) is equal to (a) ((n+1)^2)/(2n) (b) ((n+1)^2)/n (c) ((n+1)^2)/2 (d) ((n+1)^2)/(4n)