Chapter 7.5, Problem 22AYU

To determine

Find the exact value of expression.

Answer to Problem 22AYU

  $\boxed{\cot \left(-\frac{5\pi }{12}\right)=\sqrt{3}-2}$

Explanation of Solution

Given information:

Find the exact value of expression.

  $\cot \left(-\frac{5\pi }{12}\right)$

Calculation:

Consider the given repression,

  $\cot \left(-\frac{5\pi }{12}\right)$

Rewrite $\cot \left(-\frac{5\pi }{12}\right)$ as

  $\begin{array}{l}\cot \left(-\frac{5\pi }{12}\right)=\left[\therefore \cot \theta =\frac{1}{\tan \theta }\right]\\ \frac{1}{-\tan \left(\frac{5\pi }{12}\right)}=-\frac{1}{\tan \left(\frac{2\pi }{12}+\frac{3\pi }{12}\right)}\end{array}$

  $=-\frac{1}{\tan \left(\frac{\pi }{6}+\frac{\pi }{4}\right)}$

Apply identity,

  $\tan \left(\alpha +\beta \right)=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }$

  $\begin{array}{l}=-\frac{1}{\frac{\tan \frac{\pi }{6}+\tan \frac{\pi }{4}}{1-\tan \frac{\pi }{6}\tan \frac{\pi }{4}}}\\ =\frac{\tan \frac{\pi }{6}\tan \frac{\pi }{4}-1}{\tan \frac{\pi }{6}+\tan \frac{\pi }{4}}=\frac{\frac{1}{\sqrt{3}}\times 1-1}{\frac{1}{\sqrt{3}}+1}\\ =\frac{\frac{1}{\sqrt{3}}-1}{\frac{1}{\sqrt{3}}+1}=\frac{\frac{1-\sqrt{3}}{\sqrt{3}}}{\frac{1+\sqrt{3}}{\sqrt{3}}}=\frac{1-\sqrt{3}}{1+\sqrt{3}}\\ =\frac{1-\sqrt{3}}{1+\sqrt{3}}\times \frac{1-\sqrt{3}}{1-\sqrt{3}}=\frac{1-\sqrt{3}-\sqrt{3}+3}{1-3}\\ =\frac{4-2\sqrt{3}}{-2}\end{array}$

  $\cot \left(-\frac{5\pi }{12}\right)=\sqrt{3}-2$

Hence, $\boxed{\cot \left(-\frac{5\pi }{12}\right)=\sqrt{3}-2}$