$\sin {20^o}\,\sin {40^o}\,\sin {60^o}\,\sin {80^o} = $
$\sin {20^o}\,\sin {40^o}\,\sin {60^o}\,\sin {80^o} = $
- A
$ - 3/16$
- B
$5/16$
- C
$3/16$
- D
$ - 5/16$
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- [IIT 1964]
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