Chapter 14.7, Problem 16WE

To determine

To Find: The value of the given trigonometric equation.

Answer to Problem 16WE

The value of the expression is $\frac{56}{65}$ .

Explanation of Solution

Given:

The given equation is $\cos \left({\tan }^{-1}\frac{4}{3}-{\sin }^{-1}\frac{5}{13}\right)$ .

Calculation:

Consider the given trigonometric equation is,

  $\cos \left({\tan }^{-1}\frac{4}{3}-{\sin }^{-1}\frac{5}{13}\right)$

Consider $x={\tan }^{-1}\frac{4}{3}$ then,

  $\begin{array}{l}\tan x=\frac{4}{3}\\ \sec x=\sqrt{{\left(\frac{4}{3}\right)}^2+1}\\ \sec x=\frac{5}{3}\\ \cos x=\frac{3}{5}\end{array}$

Consider $\cos x=\frac{3}{5}$ then,

  $\begin{array}{l}\sin x=\sqrt{1-{\left(\frac{3}{5}\right)}^2}\\ x={\sin }^{-1}\frac{4}{5}\end{array}$

Consider that ${\sin }^{-1}\frac{5}{13}$ is,

  $\begin{array}{l}y={\sin }^{-1}\frac{5}{13}\\ \sin y=\frac{5}{13}\\ \cos y=\sqrt{1-{\left(\frac{5}{13}\right)}^2}\\ \cos y=\frac{12}{13}\end{array}$

Then, solve the given trigonometric equation as,

  $\begin{array}{c}\cos \left[{\tan }^{-1}\left(\frac{4}{3}\right)-{\sin }^{-1}\left(\frac{5}{13}\right)\right]=\cos \left[{\cos }^{-1}\left(\frac{3}{5}\right)-{\cos }^{-1}\left(\frac{12}{13}\right)\right]\\ =\cos {\cos }^{-1}\left(\frac{3}{5}\right)\cos {\cos }^{-1}\left(\frac{12}{13}\right)+\sin {\sin }^{-1}\left(\frac{4}{5}\right)\sin {\sin }^{-1}\left(\frac{5}{13}\right)\\ =\left(\frac{3}{5}\right)\left(\frac{12}{13}\right)+\left(\frac{4}{5}\right)\left(\frac{5}{13}\right)\\ =\frac{56}{65}\end{array}$