Prove that `t a n A+2t a n2A+4t a n4A+8cot8A=cot Adot`

Prove that: t a n 70^0=t a n 20^0+2t a n 50^0dot

Prove that : t a n A+tan(60^0+A)+tan(120^0+A)=3t a n3A

prove that : tan(alpha)+2 tan(2alpha) +4(tan4alpha)+8cot(8alpha) = cot(alpha)

If A+B+C=pi , prove that : cot (A/2)+ cot(B/2) + cot(C/2) = cot(A/2) cot (B/2) cot (C/2)

Let theta in (0,pi/4) and t_1=(tan theta)^(tan theta), t_2=(tan theta)^(cot theta), t_3=(cot theta)^(tan theta) and t_4=(cot theta)^(cot theta), then

if A =[(3,-4),( 1,-1)] then prove that A^n = [ ( 1+2n, -4n),( n,1-2n)] where n is any positive integer .

if A =[(3,-4),( 1,-1)] then prove that A^n = [ ( 1+2n, -4n),( n,1-2n)] where n is any positive integer .

Prove that : cot(pi/4-2cot^(-1)3)=7

The following geometric arrays suggest a sequence of numbers. (a) Find the next three terms . (b) Find the 100^(th) term . (c) Find the n^(th) term . The dots here arranged in such a way that they form a rectangle. Here T_1 =2 , T_2 = 6 , T_3 = 12 , T_4 = 20 and so on. Cay you guess what T_5 is ? What about T_6 ? What about T_n ? Make a conjecture about T_n .

The solution of the inequality (tan^(-1)x)^2-3tan^(-1)x+2geq0 is a. (-oo,\ t a n1]uu[t a n2,\ oo) b. (-oo,\ t a n1] c. (-oo,\ -t a n1]uu[t a n2,oo) d. [-1,\ 1]-\ (-(sqrt(2))/2,(sqrt(2))/2)