Chapter 13, Problem 14PT

To determine

To find:Exact value of each function.

Answer to Problem 14PT

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

Explanation of Solution

Given information:

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

Given

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

Let us assume

  ${\cos }^{-1}\frac{4}{5}=x$

On simplifying

We get

  $\begin{array}{l}{\cos }^{-1}\frac{4}{5}=x\\ \Rightarrow \cos x=\frac{4}{5}\end{array}$

We know

  $\cos x=\frac{x}{r}$

So, on comparing

We get

  $\begin{array}{l}x=4\\ r=5\end{array}$

We know

  $r=\sqrt{x^2+y^2}$

On substituting values

We get

  $\begin{array}{l}r=\sqrt{x^2+y^2}\\ \Rightarrow 5=\sqrt{4^2+y^2}\\ \Rightarrow 5=\sqrt{16+y^2}\\ \Rightarrow 5^2={\left(\sqrt{16+y^2}\right)}^2\\ \Rightarrow 25=16+y^2\\ \Rightarrow 25-16=y^2\\ \Rightarrow y^2=9\\ \Rightarrow y=3\end{array}$

Therefore,

  $\begin{array}{l}\tan x=\frac{y}{x}\\ \Rightarrow \tan x=\frac{3}{4}\\ \Rightarrow x={\tan }^{-1}\left(\frac{3}{4}\right)\end{array}$

Therefore,

  $\begin{array}{l}\tan \left({\cos }^{-1}\frac{4}{5}\right)=\tan \left(x\right)\\ \Rightarrow \tan \left({\cos }^{-1}\frac{4}{5}\right)=\tan \left({\tan }^{-1}\left(\frac{3}{4}\right)\right)\end{array}$

Cancelling tan and tan inverse

We get

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

Therefore,

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