Question

What is equal to $\tan \left( \dfrac{\pi }{12} \right)$?
A. $2-\sqrt{3}$
B. $2+\sqrt{3}$
C. $\sqrt{2}-\sqrt{3}$
D. $\sqrt{3}-\sqrt{2}$

Answer 1

Hint: To solve the question, we have to apply trigonometric identities and the values of trigonometric functions to arrive at the value of $\tan \left( \dfrac{\pi }{12} \right)$.

Complete step-by-step answer:

We know that the formula for $\tan 2\alpha$ is given by $\dfrac{2\tan \alpha }{1-{{\tan }^{2}}\alpha }$
By substituting $\alpha =\dfrac{\pi }{12}$ in the above formula we get
$\tan \left( \dfrac{2\pi }{12} \right)=\dfrac{2\tan \left( \dfrac{\pi }{12} \right)}{1-{{\left( \tan \left( \dfrac{\pi }{12} \right) \right)}^{2}}}$
$\tan \left( \dfrac{\pi }{6} \right)=\dfrac{2\tan \left( \dfrac{\pi }{12} \right)}{1-{{\left( \tan \left( \dfrac{\pi }{12} \right) \right)}^{2}}}$ …….. (1)
We know that the value of $\tan \left( \dfrac{\pi }{6} \right)$ is equal to $\dfrac{1}{\sqrt{3}}$
By substituting the above mentioned value in equation (1) we get,
$\dfrac{1}{\sqrt{3}}=\dfrac{2\tan \left( \dfrac{\pi }{12} \right)}{1-{{\left( \tan \left( \dfrac{\pi }{12} \right) \right)}^{2}}}$
Cross multiply the above expression to obtain a quadratic expression of $\tan \left( \dfrac{\pi }{12} \right)$.
$1-{{\left( \tan \left( \dfrac{\pi }{12} \right) \right)}^{2}}=\sqrt{3}\left( 2\tan \left( \dfrac{\pi }{12} \right) \right)$
${{\left( \tan \left( \dfrac{\pi }{12} \right) \right)}^{2}}+2\sqrt{3}\left( \tan \left( \dfrac{\pi }{12} \right) \right)-1=0$ …….. (2)
We know that the solutions of the general quadratic expression $a{{x}^{2}}+bx+c=0$ are given by $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
On comparing the above expression with equation (2) we get,
The values of a = 1, b = $2\sqrt{3}$, c = -1
Thus, the possible values of $\tan \left( \dfrac{\pi }{12} \right)$ are equal to $\dfrac{-2\sqrt{3}\pm \sqrt{{{\left( 2\sqrt{3} \right)}^{2}}-4(1)(-1)}}{2(1)}$
We know that ${{\left( ab \right)}^{m}}={{a}^{m}}\times {{b}^{m}}$
$=\dfrac{-2\sqrt{3}\pm \sqrt{\left( {{2}^{2}}\times 3 \right)+4}}{2}$
$=\dfrac{-2\sqrt{3}\pm \sqrt{12+4}}{2}$
$=\dfrac{-2\sqrt{3}\pm \sqrt{16}}{2}$
$=\dfrac{-2\sqrt{3}\pm 4}{2}$
Since we know that the value of $\sqrt{16}=\sqrt{4\times 4}=\sqrt{{{4}^{2}}}=4$.
$\Rightarrow \tan \left( \dfrac{\pi }{12} \right)=\dfrac{2\left( -\sqrt{3}\pm 2 \right)}{2}$
$\tan \left( \dfrac{\pi }{12} \right)=-\sqrt{3}\pm 2$
We know that $\tan \alpha$ is positive in the interval $0<\alpha <\dfrac{\pi }{2}$ . Thus, we get
$\tan \left( \dfrac{\pi }{12} \right)=2-\sqrt{3}$
$\therefore$ The value of $\tan \left( \dfrac{\pi }{12} \right)$ is equal to $2-\sqrt{3}$

Note: The possibility of mistake can be the calculation since the procedure of solving requires square root terms. The other possibility of mistake is not being able to choose the correct answer out of the obtained two values. The alternative way of solving can be, to calculate the value of $\cos \left( \dfrac{\pi }{12} \right),\sin \left( \dfrac{\pi }{12} \right)$ since we know the value of $\cos \left( \dfrac{\pi }{6} \right)$ is equal to $\dfrac{\sqrt{3}}{2}$ . By substituting the values in the formula $\cos 2\alpha =2{{\cos }^{2}}\alpha -1=1-2{{\sin }^{2}}\alpha$ , we can calculate the value of $\cos \left( \dfrac{\pi }{12} \right),\sin \left( \dfrac{\pi }{12} \right)$. Using the formula $\tan \alpha =\dfrac{\sin \alpha }{\cos \alpha }$ we can calculate the value of $\tan \left( \dfrac{\pi }{12} \right)$. This method eases the procedure of solving.