Chapter 5.4, Problem 49E

To determine

Tofind:Thevalue of the trigonometric expression $\tan \left(u-v\right)$ for $\sin u=-\frac{7}{25}$ and $\cos v=-\frac{4}{5}$ where $u$ and $v$ are in quadrant III.

Answer to Problem 49E

Thevalue of the trigonometric expression $\tan \left(u-v\right)$ is $-\frac{11}{117}$ .

Explanation of Solution

Given information:

The trigonometric expression is $\tan \left(u-v\right)$ . The value of $\sin u=-\frac{7}{25}$ and $\cos v=-\frac{4}{5}$ where $u$ and $v$ are in quadrant III.

Calculation:

Since, the value of $u$ and $v$ are in quadrant III. Let’s calculate the value of $\cos u$ and $\tan u$ in quadrant III using Pythagoras theorem.

  

Calculate the base of the triangle when $\sin u=-\frac{7}{25}$ .

  $\begin{array}{c}\text{Base}=\sqrt{{\left(25\right)}^2-{\left(-7\right)}^2}\\ =\sqrt{576}\\ =\pm 24\end{array}$

Since, the values lies in quadrant III. Take the negative value of base.

Calculate the value of $\cos u$ and $\tan u$ .

  $\begin{array}{l}\cos u=-\frac{24}{25}\\ \tan u=\frac{7}{24}\end{array}$

Similar, let’s calculate the value of $\sin v$ and $\tan v$ in quadrant III using Pythagoras theorem.

  

Calculate the base of the triangle when $\cos v=-\frac{4}{5}$ .

  $\begin{array}{c}\text{Base}=\sqrt{{\left(5\right)}^2-{\left(-4\right)}^2}\\ =\sqrt{25-16}\\ =\pm 3\end{array}$

Since, the values lies in quadrant III. Take the negative value of the base.

Calculate the value of $\sin v$ and $\tan v$ .

  $\begin{array}{c}\sin v=-\frac{3}{5}\\ \tan v=\frac{3}{4}\end{array}$

Expand the given expression using sum formula.

  $\sin \left(u+v\right)=\sin u\cdot \cos v+\cos u\cdot \sin v$

Substitute the values in the above expression.

  $\begin{array}{c}\tan \left(u-v\right)=\frac{\tan u-\tan v}{1+\tan u\tan v}\\ =\frac{\left(\frac{7}{24}-\frac{3}{4}\right)}{\left(1+\frac{7}{24}\cdot \frac{3}{4}\right)}\\ =\frac{7-18}{96+21}\\ =-\frac{11}{117}\end{array}$

Therefore, thevalue of the trigonometric expression $\tan \left(u-v\right)$ is $-\frac{11}{117}$ .