Chapter 7.3, Problem 9CFU

To determine

To find: the exact trigonometric value by using the given information.

Answer to Problem 9CFU

  $-\frac{63}{16}$

Explanation of Solution

Given information:

  $\begin{array}{l}Trignometric\text{ function: tan}\left(x+y\right)\\ \csc x=\frac{5}{3},0<x<\frac{\pi }{2}\\ \cos y=\frac{5}{13},0<y<\frac{\pi }{2}\end{array}$

Calculation:

Evaluate the value of $\sin x\text{ and }\cos x$ as shown below.

  $\begin{array}{c}\sin x=\frac{1}{\csc x}\left[\text{Reciprocal identity}\right]\\ =\frac{1}{\left(\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\right)}\left[\text{Put the value of }\csc x\right]\\ =\frac{3}{5}\\ \cos x=\sqrt{1-{\sin }^{2}x}\left[\text{Pythagorean identity}\right]\\ =\sqrt{1-{\left(\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\right)}^2}\left[\text{Put the value of }\sin x\right]\\ =\sqrt{1-\left(\raisebox{1ex}{$9$}\!\left/ \!\raisebox{-1ex}{$25$}\right.\right)}\\ =\sqrt{\raisebox{1ex}{$16$}\!\left/ \!\raisebox{-1ex}{$25$}\right.}\\ =\frac{4}{5}\end{array}$

Similarly evaluate the value of $\sin y$ by applying Pythagorean identity as shown below.

  $\begin{array}{c}\sin y=\sqrt{1-{\cos }^{2}y}\left[\text{Pythagorean identity}\right]\\ =\sqrt{1-{\left(\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$13$}\right.\right)}^2}\left[\text{Put the value of }\cos y\right]\\ =\sqrt{1-\left(\raisebox{1ex}{$25$}\!\left/ \!\raisebox{-1ex}{$169$}\right.\right)}\\ =\sqrt{\frac{144}{169}}\\ =\frac{12}{13}\end{array}$

Now evaluate $\tan x\text{ and}\tan y$ by applying quotient identity as shown below.

  $\begin{array}{c}\tan x=\frac{\sin x}{\cos x}\\ =\frac{\left(\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\right)}{\left(\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\right)}\left[\text{Put the value for }\sin x\text{ and }\cos x\right]\\ =\frac{3}{4}\\ \\ \tan y=\frac{\sin y}{\cos y}\\ =\frac{\left(\raisebox{1ex}{$12$}\!\left/ \!\raisebox{-1ex}{$13$}\right.\right)}{\left(\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$13$}\right.\right)}\left[\text{Put the value for }\sin y\text{ and }\cos y\right]\\ =\frac{12}{5}\end{array}$

Now solve $\text{tan}\left(x+y\right)$ by applying sum identity for tangent as shown below.

  $\begin{array}{c}\tan \left(x+y\right)=\frac{\tan x+\tan y}{1-\tan x\cdot \tan y}\\ =\frac{\frac{3}{4}+\frac{12}{5}}{1-\frac{3}{4}\cdot \frac{12}{5}}\left[\text{Put the values for tangent functions}\right]\\ =\frac{\frac{15+48}{20}}{\frac{20-36}{20}}\\ =\frac{15+48}{20-36}\\ =-\frac{63}{16}\end{array}$

Thus the value of given trigonometric function is $-\frac{63}{16}$ .