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Q.

tan1(47)+tan1(419)+tan1(439)+tan1(467)+......=tan1k  then  k=

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a

3

b

1

c

2

d

12

answer is C.

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Detailed Solution

tan1(47)+tan1(419)+tan1(439)+tan1(467)+......=tan1k
Sn=limn=r1ntan144r2+3
Sn=limn=r1ntan11r2+34
Sn=limn=r1ntan1r+12-r-121-r2-14
=limnr=1ntan1r+12tan1r12=tan1()-tan112=π2-tan112=cot112=tan1(2)

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