How do we find ${tan 120}^{\circ}$ $tan120^@$ using double angle identity?

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Answer 1 · 2016-11-07T06:13:07

${tan 120}^{o} = - \sqrt{3}$ $tan120^o=-sqrt3$

We use the double angle identity $tan A = \frac{2 tan A}{1 - {tan}^{2} A}$ $tanA=(2tanA)/(1-tan^2A)$ .

As such ${tan 120}^{o} = tan (2 \times 60^{o})$ $tan120^o=tan(2xx60^o)$

Hence, ${tan 120}^{o} = \frac{2 \times {tan 60}^{o}}{1 - {tan}^{2} 60^{o}}$ $tan120^o=(2xxtan60^o)/(1-tan^(2)60^o)$

= $\frac{2 \times \sqrt{3}}{1 - {(\sqrt{3})}^{2}}$ $(2xxsqrt3)/(1-(sqrt3)^2)$

= $\frac{2 \sqrt{3}}{1 - 3} = \frac{2 \sqrt{3}}{- 2} = - \sqrt{3}$ $(2sqrt3)/(1-3)=(2sqrt3)/(-2)=-sqrt3$

How do we find tan120∘tan120^@ using double angle identity?