Chapter 9.7, Problem 10E

To determine

To find: To find all trigonometric functions.

Answer to Problem 10E

  $\begin{array}{l}\sin \theta =-\frac{3}{5},\cos \theta =-\frac{4}{5},\tan \theta =\frac{3}{4}\\ \sec \theta =-\frac{5}{4},\cot \theta =\frac{4}{3}\end{array}$

Explanation of Solution

Given information:

  $\csc \theta =-\frac{5}{3},$

  $\pi <\theta <\frac{3\pi }{2}$

Formula used:

  $\begin{array}{l}\sin \theta =\frac{Perpendicular}{Hypotenuse},\cos \theta =\frac{Base}{Hypotenuse},\tan \theta =\frac{Perpendicular}{Base}\\ \csc \theta =\frac{Hypotenuse}{Perpendicular},\sec \theta =\frac{Hypotenuse}{Base},\cot \theta =\frac{Base}{Perpendicular}\end{array}$

  ${\left(Hypotenuse\right)}^2={\left(Perpendicular\right)}^2+{\left(Base\right)}^2$

  $\csc \theta =\frac{Hypotenuse}{Perpendicular}=-\frac{5}{3}$

Negative sign will be neglected, because it only denotes that the given angle is in quadrant in which the trigonometric function is negative,

By Pythagoras theorem,

  $\begin{array}{l}{\left(Hypotenuse\right)}^2={\left(Perpendicular\right)}^2+{\left(Base\right)}^2\\ \\ {\left(5\right)}^2={\left(3\right)}^2+{\left(Base\right)}^2\\ \\ 25=9+{\left(Base\right)}^2\\ \\ 16={\left(Base\right)}^2\end{array}$

  $\left(Base\right)=4$

Given that $\pi <\theta <\frac{3\pi }{2}$ , So $\tan \theta ,\cot \theta$ will be positive, all others will be negative

  $\begin{array}{l}\sin \theta =-\frac{3}{5},\cos \theta =-\frac{4}{5},\tan \theta =\frac{3}{4}\\ \sec \theta =-\frac{5}{4},\cot \theta =\frac{4}{3}\end{array}$