Find tan13π/12
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Sia ? 6 years, 6 months ago
tan(13π/12)
= tan(4π/3 - π/4),
= [tan(4π/3) - tan(π/4)]/[1 + tan(4π/3)tan(π/4)]
= (-1 + √3)/(1 + √3)
= -(1 - √3)/(1 + √3)
= -(1 - √3)^2/[(1 + √3)(1 - √3)],
= -(1 - √3)^2/(1 - 3)
= (1/2)(1 - √3)^2
= (1/2)(1 - 2√3 + 3)
= (1/2)(4 - 2√3)
= 2 - √3.
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