Chapter 6, Problem 22RE

To determine

Exact value of given expression.

Answer to Problem 22RE

  $1$

Explanation of Solution

Given information:

  $\frac{1}{{\cos }^2{40}^{°}}-\frac{1}{{\cot }^2{40}^{°}}$

Formula used:

  $\begin{array}{l}{\sec }^2\theta -{\tan }^2\theta =1\\ \frac{1}{\cos \theta }=\sec \theta \\ \frac{1}{\cot \theta }=\tan \theta \end{array}$

  $\begin{array}{l}\frac{1}{{\cos }^2{40}^{°}}-\frac{1}{{\cot }^2{40}^{°}}\\ ={\left(\frac{1}{\cos {40}^{°}}\right)}^2-{\left(\frac{1}{{\cot }^2{40}^{°}}\right)}^2\\ ={\sec }^2{40}^{°}-{\tan }^2{40}^{°}\end{array}$

As we know,

  ${\sec }^2\theta -{\tan }^2\theta =1$

  $\begin{array}{l}{\sec }^2{40}^{°}-{\tan }^2{40}^{°}\\ =1\end{array}$