Chapter 12.3, Problem 44PPS

To determine

To find: Value of $\tan \left(-\frac{7\pi }{6}\right)$ .

Answer to Problem 44PPS

  $\tan \left(-\frac{7\pi }{6}\right)=-\frac{1}{\sqrt{3}}$

Explanation of Solution

Given information:

  $\tan \left(-\frac{7\pi }{6}\right)$

Concept:

If $\theta$ is the non-quadrantal angle in standard position, its reference angle ${\theta }^{\text{'}}$ is the acute angle formed by the terminal side of $\theta$ and the x-axis.

The ruled for finding the measures of reference angles for $0^0<\theta <{360}^0$ or $0^0<\theta <2\pi$ are shown

For Quadrant − I

  ${\theta }^{\text{'}}=\theta$

For Quadrant − II

  ${\theta }^{\text{'}}={180}^0-\theta$ or ${\theta }^{\text{'}}=\pi -\theta$

For Quadrant − III

  ${\theta }^{\text{'}}=\theta -{180}^0$ or ${\theta }^{\text{'}}=\theta -\pi$

For Quadrant − IV

  ${\theta }^{\text{'}}={360}^0-\theta$ or ${\theta }^{\text{'}}=2\pi -\theta$

Given

  $\tan \left(-\frac{7\pi }{6}\right)$

Here $\theta =-\frac{7\pi }{6}$

Coterminal angle $=2\pi -\frac{7\pi }{6}=\frac{12\pi -7\pi }{6}=\frac{5\pi }{6}$

It means

The terminal side of $\frac{5\pi }{6}$ lies in Quadrant − II

Therefore,

  $\begin{array}{l}{\theta }^{\text{'}}=\pi -\theta \\ \Rightarrow {\theta }^{\text{'}}=\pi -\frac{5\pi }{6}\\ \Rightarrow {\theta }^{\text{'}}=\frac{6\pi -5\pi }{6}\\ \Rightarrow {\theta }^{\text{'}}=\frac{\pi }{6}\end{array}$

Therefore,

  $\tan \left(-\frac{7\pi }{6}\right)=-\tan \left(\frac{\pi }{6}\right)$

Because the Tan function is negative in Quadrant − II

  $\Rightarrow \tan \left(-\frac{7\pi }{6}\right)=-\frac{1}{\sqrt{3}}$

Therefore,

  $\tan \left(-\frac{7\pi }{6}\right)=-\frac{1}{\sqrt{3}}$