Given that
π 2π 3π (n-1)π
sin(x) sin[x+___] sin[x+ ___] sin[x+ ___] ..... sin[x+ _______ ]
n n n n
== sin(nπ)
-----------
n-1
2
compute
π 2π 3π (n-1)π
cot(x) + cot[x+ ___] + cot[x+ ____] + cot[x+ ____] +.....+ cot[x+_______ ]
n n n n
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