Chapter 0, Problem 10RWDT

To determine

To prove: the given equation.

Answer to Problem 10RWDT

  $\tan \left(\arcsin x\right)=\frac{x}{\sqrt{1-x^2}}$ for $0<x<1$ is proved.

Explanation of Solution

Given:

  $\tan \left(\arcsin x\right)=\frac{x}{\sqrt{1-x^2}}$ for $0<x<1$ .

Calculation:

Let $\arcsin x=\theta$ .

Then,

  $\begin{array}{l}\sin \theta =x\\ \sin \theta =\frac{x}{1}\end{array}$

Now consider the right triangle,

  

Now by using the Pythagorean Theorem in the above triangle, it follows that,

  $\begin{array}{c}A{B}^2+B{C}^2=A{C}^2\\ x^2+B{C}^2=1^2\\ B{C}^2=1-x^2\\ BC=\sqrt{1-x^2}\end{array}$

Now,

  $\begin{array}{l}\tan \theta =\frac{AB}{BC}\\ \tan \theta =\frac{x}{\sqrt{1-x^2}}\\ \theta =\arctan \left(\frac{x}{\sqrt{1-x^2}}\right)\end{array}$

Therefore,

  $\arcsin x=\arctan \left(\frac{x}{\sqrt{1-x^2}}\right)$

Hence the given equation becomes,

  $\begin{array}{c}\tan \left(\arcsin x\right)=\tan \left[\arctan \left(\frac{x}{\sqrt{1-x^2}}\right)\right]\\ =\frac{x}{\sqrt{1-x^2}}\end{array}$