Solve this tough TRIGONOMETRY QUESTION..???

Question

SORRY!! SORRY!! the Que's wrong......

It's

1/sinA.sin2A + 1/sin2A.sin3A + 1/sin3A.sin4A ......to n terms

and the answer is 1/sinA[cotA - cot(n+1)A]

Answer 1

sin(x) + sin(y) = 2sin((x + y)/2)cos((x - y)/2)

sin(nA) + sin((n + 1)A) =

2sin((2n + 1)A/2)cos(- A/2) =

2sin((2n + 1)A/2)cos(A/2)

n

∑ 1/[2sin((2k + 1)A/2)cos(A/2)] =

k=1

. . . . . . . . . . . . . . n

{1/[2ncos(A/2)]} ∑ 1/sin((2k + 1)A/2) =

. . . . . . . . . . . . . k=1

Edit:

s/b

. . . . . . . . . . . . . n

{n/[2cos(A/2)]} ∑ 1/sin((2k + 1)A/2) =

. . . . . . . . . . . . k=1

C'mon, SOMEBODY should have a better answer!