Chapter 12, Problem 66SGR

To determine

To find: Evaluate each inverse trigonometric function.

Answer to Problem 66SGR

  $\tan \left({\cos }^{-1}\frac{1}{3}\right)=2.83$

Explanation of Solution

Given information:

  $\tan \left({\cos }^{-1}\frac{1}{3}\right)$

Given

  $\tan \left({\cos }^{-1}\frac{1}{3}\right)$

Let us assume

  ${\cos }^{-1}\frac{1}{3}=\theta$

On simplifying

We get

  $\cos \theta =\frac{1}{3}$

We know

  $\cos \theta =\frac{x}{r}$

On equating

We get

  $\begin{array}{l}x=1\\ r=3\end{array}$

We know

  $r=\sqrt{x^2+y^2}$

On substituting values

We get

  $\begin{array}{l}r=\sqrt{x^2+y^2}\\ \Rightarrow 3=\sqrt{1^2+y^2}\\ \Rightarrow 3=\sqrt{1+y^2}\\ \Rightarrow 3^2={\left(\sqrt{1+y^2}\right)}^2\\ \Rightarrow 9=1+y^2\\ \Rightarrow y^2=9-1\\ \Rightarrow y^2=8\\ \Rightarrow y=\sqrt{8}\\ \Rightarrow y=2\sqrt{2}\end{array}$

We know

  $\tan \theta =\frac{y}{x}$

On substituting values

We get

  $\begin{array}{l}\Rightarrow \tan \theta =\frac{2\sqrt{2}}{1}\\ \Rightarrow \tan \theta =2\sqrt{2}\\ \Rightarrow \theta ={\tan }^{-1}2\sqrt{2}\end{array}$

Therefore,

On substituting $\theta$

We get

  $\begin{array}{l}\tan \left({\cos }^{-1}\frac{1}{3}\right)=\tan \left(\theta \right)\\ \Rightarrow \tan \left({\cos }^{-1}\frac{1}{3}\right)=\tan \left({\tan }^{-1}\left(2\sqrt{2}\right)\right)\end{array}$

Cancelling Tan and Tan inverse

We get

  $\begin{array}{l}\Rightarrow \tan \left({\cos }^{-1}\frac{1}{3}\right)=2\sqrt{2}\\ \Rightarrow \tan \left({\cos }^{-1}\frac{1}{3}\right)=2.83\end{array}$

Therefore,

  $\tan \left({\cos }^{-1}\frac{1}{3}\right)=2.83$