Chapter 7.2, Problem 57AYU

To determine

To express: The trigonometric expression $\cos \left({\tan }^{-1}u\right)$ as an algebraic expression in u .

Answer to Problem 57AYU

The expression is $\frac{1}{\sqrt{u^2+1}}$ .

Explanation of Solution

Given information:

The trigonometric expression $\cos \left({\tan }^{-1}u\right)$ .

Formula used:

Trigonometric Identities :

  $\tan \theta =\frac{\sin \theta }{\cos \theta }$

  ${\sin }^2\theta +{\cos }^2\theta =1$ .

Calculation:

Consider ,

  $\cos \left({\tan }^{-1}u\right)$

Let

  $\begin{array}{l}\theta ={\tan }^{-1}u\\ \tan \theta =u\end{array}$ (i)

where $\tan \theta$ is between $\frac{-\pi }{2}<\theta <\frac{\pi }{2}$ and $-\infty <u<\infty$ .

Now,

  $\cos \left({\tan }^{-1}u\right)=\cos \theta$ (ii)

Re-writing equation (i) we get,

  $=u=\frac{\sin \theta }{\cos \theta }$

Squaring both the sides

  $\begin{array}{l}=u^2={\left(\frac{\sin \theta }{\cos \theta }\right)}^2\\ =u^2=\frac{{\sin }^2\theta }{{\cos }^2\theta }\end{array}$

Adding 1 on both the sides

  $\begin{array}{l}=u^2+1=\frac{{\sin }^2\theta }{{\cos }^2\theta }+1\\ =u^2+1=\frac{{\sin }^2\theta +{\cos }^2\theta }{{\cos }^2\theta }\\ =u^2+1=\frac{1}{{\cos }^2\theta }\end{array}$

  $\begin{array}{l}={\cos }^2\theta =\frac{1}{u^2+1}\\ =\cos \theta =\frac{1}{\sqrt{u^2+1}}\end{array}$

Hence , the algebraic expression is $\frac{1}{\sqrt{u^2+1}}$ .