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

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

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)

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

Prove that sum_(r=0)^(n) ""^(n)C_(r).(n-r)cos((2rpi)/(n)) = - n.2^(n-1).cos^(n)'(pi)/(n) .

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)

If n is an integer, prove that, (ii) tan{(n pi)/2 + (-1)^(n) pi/4} = 1

If n is positive integer and "cos" (pi)/(2n)+"sin" (pi)/(2n) =(sqrt(n))/(2) , then prove that 4 le n le 8

Evaluate : underset(n to oo)lim(1)/(n)["tan"(pi)/(4n)+"tan"(2pi)/(4n)+"tan"(3pi)/(4n)+…+ "tan"(npi)/(4n)]

Evaluate : underset(n to oo)lim (1)/(n)[sin (pi/(2n))+sin((2pi)/(2n))+sin((3pi)/(2n))+…+ sin((npi)/(2n))]