Chapter 6.3, Problem 47E

To determine

To calculate: The value of trigonometric ratios if $\tan \theta =-\frac{3}{4}$ and $\cos \theta >0$ .

Answer to Problem 47E

The value of trigonometric ratios are,

  $\sin \theta =-\frac{3}{5},cos\theta =\frac{4}{5},\tan \theta =-\frac{3}{4},\csc \theta =-\frac{5}{3},\sec \theta =\frac{5}{4}\text{ and }\cot \theta =-\frac{4}{3}$ .

Explanation of Solution

Given information:

The value of trigonometric function $\tan \theta =-\frac{3}{4}$ and $\cos \theta >0$ .

Formula used:

The trigonometric ratios for a right angle triangle are defined as,

  $\sin \theta =\frac{\text{opposite}}{\text{hypotenuse}},cos\theta =\frac{\text{adjacent}}{\text{hypotenuse}},\tan \theta =\frac{\text{opposite}}{\text{adjacent}},\csc \theta =\frac{\text{hypotenuse}}{\text{opposite}},\sec \theta =\frac{\text{hypotenuse}}{\text{adjacent}}$ and $\cot \theta =\frac{\text{adjacent}}{\text{opposite}}$ .

Coordinate plane is divided into four quadrants.

In the first quadrant all trigonometric functions that is $\sin \theta ,cos\theta ,\tan \theta ,\csc \theta ,\sec \theta \text{ and }\cot \theta$ are positive.

In the second quadrant only sine and cosecant trigonometric functions that is $\sin \theta \text{ and }\csc \theta$ are positive and remaining $cos\theta ,\tan \theta ,\sec \theta \text{ and }\cot \theta$ are negative.

In the third quadrant only tangent and cotangent trigonometric functions that is $\tan \theta \text{ and }\cot \theta$ are positive and remaining $cos\theta ,sin\theta ,\sec \theta \text{ and }\csc \theta$ are negative.

In the fourth quadrant only cosine and secant trigonometric functions that is $\cos \theta \text{ and }\sec \theta$ are positive and remaining $\sin \theta ,\tan \theta ,\csc \theta \text{ and }\cot \theta$ are negative.

Calculation:

Consider the provided value of trigonometric function $\tan \theta =-\frac{3}{4}$ and $\cos \theta >0$ .

Since, the tangent function is expressed as $\tan \theta =\frac{\text{opposite}}{\text{adjacent}}$ . Construct a right angle triangle with acute angle $\theta$ and whose opposite side is 3 units and adjacent side is 4 units.

The figure obtained is provided below,

Observe that opposite side is 3 units and adjacent side is 4 units.

Now, let the length of hypotenuse be x , as it is a right angle triangle so,

  $\begin{array}{c}{4}^2+3^2=x^2\\ x^2=16+9\\ x^2=25\\ x=5\end{array}$

Therefore, length of hypotenuse is $5$ units.

Recall that the trigonometric ratios for a right angle triangle are defined as,

  $\sin \theta =\frac{\text{opposite}}{\text{hypotenuse}},cos\theta =\frac{\text{adjacent}}{\text{hypotenuse}},\tan \theta =\frac{\text{opposite}}{\text{adjacent}},\csc \theta =\frac{\text{hypotenuse}}{\text{opposite}},\sec \theta =\frac{\text{hypotenuse}}{\text{adjacent}}$ and $\cot \theta =\frac{\text{adjacent}}{\text{opposite}}$ .

It is provided that $\cos \theta >0$ and $\tan \theta <0$ .

In the fourth quadrant only cosine and secant trigonometric functions that is $\cos \theta \text{ and }\sec \theta$ are positive and remaining $\sin \theta ,\tan \theta ,\csc \theta \text{ and }\cot \theta$ are negative.

Apply it, to estimate the value of trigonometric ratios,

The value of sine function is,

  $\begin{array}{c}\sin \theta =\frac{\text{opposite}}{\text{hypotenuse}}\\ =-\frac{3}{5}\end{array}$

The value of cosine function is,

  $\begin{array}{c}cos\theta =\frac{\text{adjacent}}{\text{hypotenuse}}\\ =\frac{4}{5}\end{array}$

The value of tangent function is,

  $\begin{array}{c}\tan \theta =\frac{\text{opposite}}{\text{adjacent}}\\ =-\frac{3}{4}\end{array}$

The value of cosecant function is,

  $\begin{array}{c}\csc \theta =\frac{\text{hypotenuse}}{\text{opposite}}\\ =-\frac{5}{3}\end{array}$

The value of secant function is,

  $\begin{array}{c}\sec \theta =\frac{\text{hypotenuse}}{\text{adjacent}}\\ =\frac{5}{4}\end{array}$

The value of cotangent function is,

  $\begin{array}{c}\cot \theta =\frac{\text{adjacent}}{\text{opposite}}\\ =-\frac{4}{3}\end{array}$

Hence, the value of trigonometric ratios are,

  $\sin \theta =-\frac{3}{5},cos\theta =\frac{4}{5},\tan \theta =-\frac{3}{4},\csc \theta =-\frac{5}{3},\sec \theta =\frac{5}{4}\text{ and }\cot \theta =-\frac{4}{3}$ .