Chapter 6, Problem 41RE

To determine

Value of given trigonometric expression

Answer to Problem 41RE

  $\cot \left(-{390}^{°}\right)=-\sqrt{3}$

Explanation of Solution

Given information:

  $\cot \left(-{390}^{°}\right)$

Formula used:

  $\begin{array}{l}\cot \left(-\theta \right)=-\cot \theta \\ \cot \left(k\left({180}^{°}\right)+\theta \right)=\cot \theta \end{array}$

  $\begin{array}{l}\cot \left(-\theta \right)=-\cot \theta \\ \cot \left(-{390}^{°}\right)=-\cot {390}^{°}\end{array}$

  $\begin{array}{l}-\cot {390}^{°}\\ =-\cot \left({360}^{°}+{30}^{°}\right)\end{array}$

As,

  $\begin{array}{l}\cot \left(k\left({180}^{°}\right)+\theta \right)=\cot \theta \\ -\cot \left({360}^{°}+{30}^{°}\right)=-\cot {30}^{°}\end{array}$

  $\boxed{\cot \left(-{390}^{°}\right)=-\cot \left({30}^{°}\right)=-\sqrt{3}}$