Question

How do you evaluate $\tan \left( \dfrac{11\pi }{12} \right)$ ?
(a) Using trigonometric angle identities
(b) Using linear formulas
(c) a and b both
(d) none of the above

Answer 1

Hint: In this problem we are to find the value of $\tan \left( \dfrac{11\pi }{12} \right)$. We will try to use the trigonometric double angle identities of $\tan 2\theta $ to find and simplify the value of our needed problem. We can start with the fact that $\tan \left( \pi -\theta \right)=-\tan \theta $ and consider $\theta =\dfrac{\pi }{12}$ to get ahead with the problem and evaluate the value.

Complete step by step solution:
According to the question, we start with,
$\tan \left( \dfrac{11\pi }{12} \right)$
now, $\dfrac{11\pi }{12}$ can be written as, $\pi -\dfrac{\pi }{12}$,
Thus we are getting,
$\Rightarrow \tan \left( \pi -\dfrac{\pi }{12} \right)$
As, $\tan (\pi -\theta )=-\tan \theta $ , because it is residing in the second quadrant, we get from the last line,
$\Rightarrow -\tan \left( \dfrac{\pi }{12} \right)$
Now, let us consider, $\theta =\dfrac{\pi }{12}$ , then,$2\theta =\dfrac{\pi }{6}$ ,
Then we go on with,
$\tan 2\theta =\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta }$
So, $\tan \left( \dfrac{\pi }{6} \right)=\dfrac{2t}{1-{{t}^{2}}}$ , where \[t=\tan \theta =\tan \left( \dfrac{\pi }{12} \right)\]
Putting the value of the tangent function from the trigonometric table, we are getting,
$\Rightarrow \dfrac{1}{\sqrt{3}}=\dfrac{2t}{1-{{t}^{2}}}$
Cross multiplying,
$\Rightarrow 1-{{t}^{2}}=2\sqrt{3}t$
Again, simplifying,
$\Rightarrow {{t}^{2}}+2\sqrt{3}t=1$
Adding 3 on both sides,
$\Rightarrow {{t}^{2}}+2\sqrt{3}t+{{\left( \sqrt{3} \right)}^{2}}=4$
Going ahead for more simplification,
$\Rightarrow {{\left( t+\sqrt{3} \right)}^{2}}={{2}^{2}}$
Removing the squares from both sides, we get,
$\Rightarrow t+\sqrt{3}=\pm 2$ , which says the value on the right hand side can both be positive and negative.
Then, $t=\pm 2-\sqrt{3}$
Now, taking the negative value, $t=-2-\sqrt{3}=\tan \left( \dfrac{\pi }{12} \right)$ which is not possible as, $\dfrac{\pi }{12}$ lies in the first quadrant where the value of tan is always positive.
So, again we have to take the positive value as our answer, $t=2-\sqrt{3}=\tan \left( \dfrac{\pi }{12} \right)$,
Then,
$\tan \left( \dfrac{11\pi }{12} \right)=-\tan \left( \dfrac{\pi }{12} \right)=\sqrt{3}-2$

So, the correct answer is “Option a”.

Note: To understand how the values of trigonometric ratios like $\tan \left( \dfrac{\pi }{12} \right)$ change in different quadrants, first we have to understand ASTC rule. The ASTC rule is nothing but the "all sin tan cos" rule in trigonometry. The angles which lie between 0° and 90° are said to lie in the first quadrant. The angles between 90° and 180° are in the second quadrant, angles between 180° and 270° are in the third quadrant and angles between 270° and 360° are in the fourth quadrant. In the first quadrant, the values for sin, cos and tan are positive. In the second quadrant, the values for sin are positive only. In the third quadrant, the values for tan are positive only. In the fourth quadrant, the values for cos are positive only.