ChallengeMediumTelescoping & Recursion ShareTelescoping & Recursion · Medium problemEvaluate ∑k=1891sin(k∘)sin((k+1)∘)\displaystyle\sum_{k=1}^{89}\frac{1}{\sin(k^\circ)\sin\big((k+1)^\circ\big)}k=1∑89sin(k∘)sin((k+1)∘)1.Acos1∘sin21∘\dfrac{\cos 1^\circ}{\sin^2 1^\circ}sin21∘cos1∘Bcos1∘sin22∘\dfrac{\cos 1^\circ}{\sin^2 2^\circ}sin22∘cos1∘Csin21∘cos1∘\dfrac{\sin^2 1^\circ}{\cos 1^\circ}cos1∘sin21∘Dcos1∘sin31∘\dfrac{\cos 1^\circ}{\sin^{3} 1^\circ}sin31∘cos1∘Check Show answer Walk through it step by step Practice more like this Review this concept