Find the value of tan15º without using calculator or table
1 Answer
Here,
tan15º
= tan (60º - 45º)
= \(\frac{\tan60º\;-\;\tan45º}{1+\tan60º.\tan45º}\)
= \(\frac{\sqrt3-1}{1+\sqrt3\times1}\)
= \(\frac{\sqrt3-1}{\sqrt3+1}\times\frac{\sqrt3-1}{\sqrt3-1}\)
= \(\frac{\left(\sqrt3-1\right)}{\left(\sqrt3\right)^2-1^2}^2\)
= \(\frac{3-2\sqrt3+1}{3-1}\)
= \(\frac{4-2\sqrt3}2\)
= \(\frac{2(2-\sqrt3)}2\)
= \(2-\sqrt3\)
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