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(1+sec2A)(1+sec4A)(1+sec8A)=Tan8A.CotA prove Sin6.Sin42.Sin66.Sin78=1/16

(1+sec2A)(1+sec4A)(1+sec8A)=Tan8A.CotA prove
Sin6.Sin42.Sin66.Sin78=1/16

Grade:10

1 Answers

Arun
25763 Points
3 years ago
Dear student
 
Please ask one question only in one thread.
We use the trigonometric rules :
2 Sin A  Sin B = Cos (A-B) - Cos (A+B)
2 Cos A Cos B = Cos (A-B) + Cos (A+B)
Sin 42°  Sin 78°  = 1/2 * [ Cos (42° - 78°) - Cos (78° + 42°) ]
= 1/2 * [ Cos 36° - Cos 120°] 
= 1/2 [ Cos 36° + 1/2 ]
Sin 6°  Sin 66°  = 1/2 * [ Cos (6-66)  - Cos (6+66) ]
= 1/2 [ Cos 60° - Cos 72° ]
= 1/2 [ 1/2 - Cos 72° ]
Hence,  Sin 6°  Sin 66° Sin 42°  Sin 78° 
= 1/16 [ 2 Cos 36° + 1 ] [ 1 - 2 Cos 72° ]
=  1/16 [ 2 Cos 36° - 2 cos 72° + 1 - 4 Cos 36° Cos 72°  ]
= 1/16 + 1/8 [ Cos 36° - Cos 72° - 2 Cos 36° Cos 72° ]
= 1/16 + 1/8 [ Cos 36° - Cos 72° - ( Cos (36+72) + Cos (72-36) ) ]
= 1/16  - 1/8 [ Cos 72° + Cos 108° ]
= 1/16 - 1/8 [  Cos 72° - Cos (180° - 108°) ]
= 1/16 - 1/8 [ Cos 72°  - Cos 72°  ]
= 1/16
Hence Proved.
 
Regards
Arun (askIITians forum expert)

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