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JEE Advanced 2016 Paper 2 Mathematics Q3
Mathematics
Problem
The value of
∑
k
=
1
13
1
sin
(
π
4
+
(
k
−
1
)
π
6
)
sin
(
π
4
+
k
π
6
)
\displaystyle\sum_{k=1}^{13}\frac{1}{\sin\left(\frac\pi4+\frac{(k-1)\pi}{6}\right)\sin\left(\frac\pi4+\frac{k\pi}{6}\right)}
k
=
1
∑
13
sin
(
4
π
+
6
(
k
−
1
)
π
)
sin
(
4
π
+
6
k
π
)
1
is equal to
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