What is the value of $\frac{2 tan (\frac{π}{12})}{1 - {tan}^{2} (\frac{π}{12})}$ $(2tan(pi/12))/(1-tan^2(pi/12))$ ?

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What is the value of $\frac{2 tan (\frac{π}{12})}{1 - {tan}^{2} (\frac{π}{12})}$ $(2tan(pi/12))/(1-tan^2(pi/12))$ ?

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Answer 1 · 2016-11-28T09:53:57

$\frac{2 tan (\frac{π}{12})}{1 - {tan}^{2} (\frac{π}{12})} = \frac{1}{\sqrt{3}}$ $(2tan(pi/12))/(1-tan^2(pi/12))=1/sqrt3$

Explanation:

As $tan (A + B) = \frac{tan A + tan B}{1 - tan A tan B}$ $tan(A+B)=(tanA+tanB)/(1-tanAtanB)$

if $A = B$ $A=B$ , we have $tan 2 A = \frac{tan A + tan A}{1 - tan A tan A}$ $tan2A=(tanA+tanA)/(1-tanAtanA)$

or $tan 2 A = \frac{2 tan A}{1 - {tan}^{2} A}$ $tan2A=(2tanA)/(1-tan^2A)$

Now putting $A = \frac{π}{12}$ $A=pi/12$

we have $\frac{2 tan (\frac{π}{12})}{1 - {tan}^{2} (\frac{π}{12})} = tan (2 \times \frac{π}{12}) = tan (\frac{π}{6}) = \frac{1}{\sqrt{3}}$ $(2tan(pi/12))/(1-tan^2(pi/12))=tan(2xxpi/12)=tan(pi/6)=1/sqrt3$

Answer 2 · 2016-11-28T09:55:53

Using identity $tan 2 θ = \frac{2 tan θ}{1 - {tan}^{2} θ}$ $tan2theta=(2tantheta)/(1-tan^2theta)$

we have the Expression

$\frac{2 tan (\frac{π}{12})}{1 - {tan}^{2} (\frac{π}{12})}$ $(2tan(pi/12))/ (1-tan^2(pi/12) )$

$= tan (2 \times \frac{π}{12}) = tan (\frac{π}{6}) = \frac{1}{\sqrt{3}}$ $=tan(2xxpi/12)=tan(pi/6)=1/sqrt3$

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What is the value of $\frac{2 tan (\frac{π}{12})}{1 - {tan}^{2} (\frac{π}{12})}$ $(2tan(pi/12))/(1-tan^2(pi/12))$ ?

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Explanation:

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