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The sum #1/(sin46sin47)+1/(sin47sin48)+1/(sin48sin49)+cdots+1/(sin133sin134)=# ?

1 Answer
Jul 29, 2018

#1/(sin(r)sin(r+1))=1/sin1[sin1/(sin(r)sin(r+1))]#

#=1/sin1[ sin((r+1)-r)/(sin(r)sin(r+1))]#

#=1/sin1[ (sin(r+1)cos(r)-cos(r+1)sin(r))/(sin(r)sin(r+1))]#

#=1/sin1[ (sin(r+1)cos(r))/(sin(r)sin(r+1))-(cos(r+1)sin(r))/(sin(r)sin(r+1))]#

#=1/sin1[ cot(r)-cot(r+1)]#

Now

#LHS=sum_(r=46)^(r=133)1/(sin(r)sin(r+1))#

#=1/sin1sum_(r=46)^(r=133)[ cot(r)-cot(r+1)]#

#=1/sin1(cot46-cot134)#

#=1/sin1(cot46-cot(180-46))#

#=1/sin1(cot46+cot46)#

#=2cot46csc1#