The value of ∑k=1131sin(π4+(k−1)π6)sin(π4+kπ6)\displaystyle\sum_{k=1}^{13}\frac{1}{\sin\left(\frac{\pi}{4}+\frac{(k-1)\pi}{6}\right)\sin\left(\frac{\pi}{4}+\frac{k\pi}{6}\right)}k=1∑13sin(4π+6(k−1)π)sin(4π+6kπ)1 is equal to
Official answer: C
Telescope. With Ak=π4+(k−1)π6A_k=\frac\pi4+\frac{(k-1)\pi}6Ak=4π+6(k−1)π, consecutive angles differ by π6\frac\pi66π:1sinAksinAk+1=sin(Ak+1−Ak)sinπ6 sinAksinAk+1=2(cotAk−cotAk+1).\frac1{\sin A_k\sin A_{k+1}}=\frac{\sin(A_{k+1}-A_k)}{\sin\frac\pi6\,\sin A_k\sin A_{k+1}}=2(\cot A_k-\cot A_{k+1}).sinAksinAk+11=sin6πsinAksinAk+1sin(Ak+1−Ak)=2(cotAk−cotAk+1).
Sum.2(cotπ4−cot(π4+13π6))=2(1−cot5π12)=2(1−(2−3))=2(3−1).2\left(\cot\frac\pi4-\cot\left(\frac\pi4+\frac{13\pi}6\right)\right)=2\left(1-\cot\frac{5\pi}{12}\right)=2\left(1-(2-\sqrt3)\right)=2(\sqrt3-1).2(cot4π−cot(4π+613π))=2(1−cot125π)=2(1−(2−3))=2(3−1).
(C)