Chapter 7, Problem 40RE

To determine

The simplified value of the expression $\tan \left({\csc }^{-1}u\right)$ in terms of $u$ .

Answer to Problem 40RE

The simplified value of the expression $\tan \left({\csc }^{-1}u\right)$ in terms of is $\frac{1}{\sqrt{1-u^2}}$ .

Explanation of Solution

Given information:

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

Formula used:

For the sine function and its inverse, the following holds:

  $\begin{array}{c}f\left(f^{-1}\left(x\right)\right)=\sin \left({\sin }^{-1}\left(x\right)\right)\\ =x\end{array}$

Where, $-1\le x\le 1$

Inverse of sine and inverse of cosec related as,

  ${\sin }^{-1}\left(\frac{1}{x}\right)={\csc }^{-1}x$

Where, $\left|x\right|\ge 1$ .

Tangent function and sine function related as,

  $\tan \theta =\frac{\sin \theta }{\sqrt{1-{\sin }^2\theta }}$

Calculation:

Consider the expression $\tan \left({\csc }^{-1}u\right)$ .

Recall that inverse of sine and inverse of cosec related as,

  ${\sin }^{-1}\left(\frac{1}{x}\right)={\csc }^{-1}x$

Where, $\left|x\right|\ge 1$ .

Applied it,

  $\tan \left({\csc }^{-1}u\right)=\tan \left({\sin }^{-1}\frac{1}{u}\right)$

Where, $\left|u\right|\ge 1$ .

Again recall that tangent function and sine function related as,

  $\tan \theta =\frac{\sin \theta }{\sqrt{1-{\sin }^2\theta }}$

Using it,

  $\tan \left({\sin }^{-1}\frac{1}{u}\right)=\frac{\sin \left({\sin }^{-1}\frac{1}{u}\right)}{\sqrt{1-{\left[\sin \left({\sin }^{-1}\frac{1}{u}\right)\right]}^2}}$

Where, $\left|u\right|\ge 1$ .

Observe that the value $\frac{1}{u}$ is lies between $-1$ and $1$ .

Recall that for the sine function and its inverse, the following holds:

  $\begin{array}{c}f\left(f^{-1}\left(x\right)\right)=\sin \left({\sin }^{-1}\left(x\right)\right)\\ =x\end{array}$

Where, $-1\le x\le 1$

Simplify further,

  $\begin{array}{c}\frac{\sin \left({\sin }^{-1}\frac{1}{u}\right)}{\sqrt{1-{\left[\sin \left({\sin }^{-1}\frac{1}{u}\right)\right]}^2}}=\frac{\frac{1}{u}}{\sqrt{1-{\left(\frac{1}{u}\right)}^2}}\\ =\frac{1}{\sqrt{1-u^2}}\end{array}$

Thus, the simplified value of the expression $\tan \left({\csc }^{-1}u\right)$ is $\frac{1}{\sqrt{1-u^2}}$ .