Chapter 14.7, Problem 25WE

To determine

To Find: The simplified expression for the given expression without the use of trigonometric or inverse trigonometric functions.

Answer to Problem 25WE

The required expression is $\frac{xy-1}{x+y}$ .

Explanation of Solution

Given:

The given expression is $\tan \left({\cot }^{-1}x+{\cot }^{-1}y\right)$ .

Calculation:

Consider the given expression is,

  $\tan \left({\cot }^{-1}x+{\cot }^{-1}y\right)$

Consider $a={\cot }^{-1}x$ and $b={\cot }^{-1}y$ , then,

  $\begin{array}{l}\cot a=x\\ \cot b=y\end{array}$

Then,

  $\begin{array}{l}\tan a=\frac{1}{x}\\ \tan b=\frac{1}{y}\end{array}$

Consider the trigonometric identity is,

  $\tan \left(a+b\right)=\frac{\tan a+\tan b}{1-\tan a\tan b}$

Then,

  $\begin{array}{c}\tan \left(a+b\right)=\frac{\frac{1}{x}+\frac{1}{y}}{1-\frac{1}{x}\cdot \frac{1}{y}}\\ =\frac{x+y}{xy-1}\end{array}$

Inverse the above function as,

  $\cot \left(a+b\right)=\frac{xy-1}{x+y}$

Substitute $a={\cot }^{-1}x$ and $b={\cot }^{-1}y$ in the above equation as,

  $\cot \left({\cot }^{-1}x+{\cot }^{-1}y\right)=\frac{xy-1}{x+y}$