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How to find exact value of tan 75°

User Imhvost
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tan \alpha = (sin \alpha )/(cos \alpha ) \\--------\\\\tan75^0= \frac{\big{sin75^0}}{\big{cos75^0}} \\\\sin75^0=sin(45^0+30^0)=sin45^0\cdot cos30^0+sin30^0\cdot cos45^0=\\\\.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = ( √(2) )/(2) \cdot ( √(3) )/(2) + (1)/(2) \cdot( √(2) )/(2) = ( √(6)+ √(2) )/(4)
User MEC
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