Prove this tough TRIGONOMETRY QUESTION..???

Question

Prove that the sum of

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

IS

cosecA[cotA - cot(n+1)A]

Please help it's Urgent!!!

hint: Multiply and Divide by SinA

Answer 1

if it is not a theory problem you cud do it by satisfying n=1,2,3...

Answer 2

we know sin A = sin ((k+1) A - k A) = sin (k+1) A cos k A -  sin k A cos (k+1) A

so sin A/( sin k A sin (k+1) A = cot kA - cot (k+1) A

adding them from k= 1 to n

sum sin A(/sinA.sin2A + 1/sin2A.sin3A + 1/sin3A.sin4A ......to n terms) = cot A - cot(n+1) A

or multiply by cosec A

sinA.sin2A + 1/sin2A.sin3A + 1/sin3A.sin4A ......to n terms) = cosec A(cot A - cot(n+1) A)