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a
tan520
b
tan30
c
tan40
d
tan50
answer is A.
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Detailed Solution
We have sinα+sinα+d+.....sinα+n−1dcosα+cosα+d+.....cosα+n−1d=sinα+α+n−1d2cosα+α+n−1d2∴sin1∘+sin2∘+sin3∘+sin4∘cos1∘+cos2∘+cos3∘+cos4∘=sin1+42ocos1+42o=tan52o