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sin 10^(@) sin50^(@)sin 60^(@) sin 70^(@...

`sin 10^(@) sin50^(@)sin 60^(@) sin 70^(@)=m " and tan "20^(@) tan 40^(@) tan60^(@) tan80^(@)= n,` then `(n)/(m)=`

A

`(3sqrt(3))/(16)`

B

`16sqrt(3)`

C

`(16)/sqrt(3)`

D

`8sqrt(3)`

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The correct Answer is:
B
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tan 20^(@) tan 40^(@) tan 60^(@) tan 80^@=

tan 70^(@) - tan 20^(@)=

sin 20^(@) * sin 40^(@) * sin 60^@ * sin 80^(@) =

tan10^(@) tan 20^(@) tan 30^(@) tan 40^(@) tan 50^(@) tan 60^(@) tan 70^(@) tan 80^(@)=

sin 10^(@) + sin 20^(@) + sin 40^(@) + sin 50^(@) - sin 70^(@) - sin 80^(@) =

(tan80^(@)-tan10^(@))/(tan70^(@))

Prove that tan 50^(@) -tan40^(@)-2tan10^(@) .

A =(cos 9^(@) - sin 9^(@))/(cos 9^(@) + sin 9^(@)) , B = (cos 21^(@) + sin 21^(@))/(cos21^(@) -sin21^(@)) and C= tan 20^(@) + tan40^(@) + sqrt(3)tan20^(@) tan40^(@) then descending order is:

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