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Prove that:`tan6^0tan42^0tan66^0tan78^0=1.`

Answer» `L.H.S. = tan6^@tan42^@tan66^@tan78^@`
`=(sin6^@sin42^@sin66^@sin78^@)/(cos6^@cos42^@cos66^@cos78^@)`
`=((2sin6^@sin66^@)(2sin42^@sin78^@))/((2cos6^@cos66^@)(2cos42^@cos78^@))`
`=((cos60^@-cos72^@)(cos36^@-cos120^@))/((cos60^@+cos72^@)(cos36^@+cos120^@))`
`=((1/2-sin18^@)(cos36^@+1/2))/((1/2+sin18^@)(cos36^@-1/2))...[As costheta = sin(90-theta)]`
`=((1-2sin18^@)(2cos36^@+1))/((1+2sin18^@)(2cos36^@-1))`
Now, putting, `sin18^@ = (sqrt5-1)/4 and cos36^@ = (sqrt5+1)/4`
`=((1-2((sqrt5-1)/4))((2(sqrt5+1)/4)+1))/((1+2((sqrt5+1)/4))(2((sqrt5-1)/4)-1))`
`=((3-sqrt5)(sqrt5+3))/((1+sqrt5)(sqrt5-1))`
`=(9-5)/(5-1)`
`=4/4 =1 = R.H.S.`


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