Chapter 7.1, Problem 59E

To determine

To state: the segment that represent the measurement of all trigonometric ratio.

Answer to Problem 59E

  $EB=\cos ec\theta -1$ and $BC=\sec \theta -\cos ec\theta$ ..

Explanation of Solution

Given:

  

Concept used:

Trigonometry formula:

  $\begin{array}{l}\sin \left(-\theta \right)=-\sin \theta .\\ \cos \left(-\theta \right)=\cos \theta .\\ \tan \left(-\theta \right)=-\tan \theta .\end{array}$

Calculation:

Consider the unit circle with center at origin and $\overleftrightarrow{AB}\text{ and }\overleftrightarrow{CD}$ are tangent to the circle.

  

From $\Delta OFE$ ;

  $\begin{array}{l}\sin \theta =\frac{\text{opposite side}}{\text{hypotenuse}}.\\ \Rightarrow \frac{EF}{1}.\\ \Rightarrow EF.\end{array}$

From $\Delta OAB$ ,

  $\begin{array}{l}\sin \theta =\frac{\text{opposite side}}{\text{hypotenuse}}.\\ \Rightarrow \frac{OA}{OB}.\\ \sin \theta =\frac{1}{OB}.\end{array}$

Taking reciprocal on both the side.

  $\begin{array}{l}\frac{1}{\sin \theta }=OB.\\ \cos ec\theta =OB.\end{array}$

From $\Delta ODC$ ,

  $\begin{array}{l}\cos \theta =\frac{\text{adjacent side}}{\text{hypotenuse}}.\\ \Rightarrow \frac{OD}{OC}.\\ \cos \theta =\frac{1}{OC}.\end{array}$

Taking reciprocal on both sides:

  $\begin{array}{l}\frac{1}{\cos \theta }=OC.\\ \sec \theta =OC.\end{array}$

From figure, in $\Delta OFE$ :

  $\begin{array}{l}\cos \theta =\frac{\text{adjacent side}}{\text{hypotenuse}}.\\ \Rightarrow \frac{OF}{1}.\\ \Rightarrow OF.\end{array}$

From $\Delta OAB$ ,

  $\begin{array}{l}\tan A=\frac{\text{opposite side}}{\text{adjacent}}\text{.}\\ \Rightarrow \frac{OA}{AB}.\\ \Rightarrow \frac{1}{AB}.\end{array}$

Taking reciprocal on both sides:

  $\begin{array}{l}\frac{1}{\tan \theta }=AB.\\ \cot \theta =AB.\end{array}$

From $\Delta ODC$ :

  $\begin{array}{l}\tan \theta =\frac{\text{opposite side}}{\text{adjacent}}\text{.}\\ \Rightarrow \frac{CD}{OD}.\\ \Rightarrow CD.\end{array}$

  $\begin{array}{l}EB=OB-OE.\\ EB=\cos ec\theta -1.\\ \text{and}\\ BC=OC-OB.\\ BC=\sec \theta -\cos ec\theta .\end{array}$

Hence, $EB=\cos ec\theta -1$ and $BC=\sec \theta -\cos ec\theta$ .