1.

Prove that `(2cos2A+1)/(2cos2A-1)=tan(60^0+A)tan(60^0-A)dot`

Answer» RHS `=tan(60^(@)+A)tan(60^(@)-A)`
`=((tan60^(@)+tanA)/(1-tan60^(@)tana))((tan60^(@)-tanA)/(1+tan60^(@)tanA)) = ((sqrt(3)+tanA)/(1-sqrt(3)tanA))((sqrt(3)-tanA)/(1+sqrt(3)tanA))`
`=(3-tan^(2)A)/(1-3tan^(2)A) = (3-(sin^(2)A)(cos^(2)A))/(1-3(sin^(2)A)/(cos^(2)A)) = (3cos^(2)A-sin^(2)A)/(cos^(2)A-3sin^(2)A) = (2cos^(2)A+cos^(2)A-2sin^(2)A+sin^(2)A)/(2cos^(2)A-2sin^(2)A-sin^(2)A-cos^(2)A)`
`=(2(cos^(2)A-sin^(2)A)+cos^(2)A+sin^(2)A)/(2(cos^(2)A-sin^(2)A)-cos^(2)A)`
`=(2(cos^(2)A-sin^(2)A)+cos^(2)A+sin^(2)A)/(2(cos^(2)A-sin^(2)A))-(sin^(2)A+cos^(2)A) = (2cos2A+1)/(2cos2A-1)`= LHS


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