Chapter 7, Problem 23SGA

To determine

To find: the exact value of given function by using sum or difference identity.

Answer to Problem 23SGA

  $\sqrt{3-2}$

Explanation of Solution

Given information:

  $\tan \left(\frac{11\pi }{12}\right)$

Calculation:

Solve the given trigonometric function as follows:

  $\begin{array}{c}\tan \left(\frac{11\pi }{12}\right)=\tan \left(\frac{\pi }{4}+\frac{2\pi }{3}\right)\left[\text{Rewrite }\frac{11\pi }{12}\right]\end{array}$

  $\begin{array}{c}=\frac{\tan \frac{\pi }{4}+\tan \frac{2\pi }{3}}{1-\tan \frac{\pi }{4}\tan \frac{2\pi }{3}}\left[\text{Use sum identity of tangent}\right]\end{array}$

  $\begin{array}{c}=\frac{1+\left(-\sqrt{3}\right)}{1-1\left(-\sqrt{3}\right)}\end{array}$

  $\begin{array}{c}=\frac{1-\sqrt{3}}{1+\sqrt{3}}\end{array}$

  $\begin{array}{c}=\frac{1-\sqrt{3}}{1+\sqrt{3}}\left(\frac{1-\sqrt{3}}{1-\sqrt{3}}\right)\left[\text{Multiply and divide by }\left(1-\sqrt{3}\right)\right]\end{array}$

  $\begin{array}{c}=\frac{4-2\sqrt{3}}{-2}\end{array}$

  $\begin{array}{c}=\sqrt{3-2}\end{array}$

Thus the value of $\tan \left(\frac{11\pi }{12}\right)$ is $\sqrt{3-2}$ .