If cos π 4 cos π 8 cos π 16 cos π 32 = 2 m cosec π n then m + n =

If cosπ4cosπ8cosπ16cosπ32=2mcosecπn then m+n=
  1. 27
  2. 25
  3. 28
  4. 29

Solution

cosπ4cosπ8cosπ16cosπ32=2mcosecπn

LHS=12sinπ2n2cosπ2nsinπ2ncosπ2n-1..........cosπ4

12cosπ2n2sinπ2n-1cosπ2n-1cosπ2n-2..........cosπ4

12×22cosecπ2n2sinπ2n-2cosπ2n-2cosπ2n-2..........cosπ4

12n-2cosecπ2n2sinπ8cosπ8cosπ4

12×2n-2cosecπ2n2sinπ4cosπ4

12n-1cosecπ2nsinπ2

12n-1cosecπ2n

if n=5 m=23,   m+n=28

Asked in: AP EAMCET 2021 (20 Aug Shift 2)

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