Chapter 5.2, Problem 19E

To determine

To verify the identity $\frac{1}{\tan x}+\frac{1}{\cot x}=\tan x+\cot x$ algebraically and to use table feature of graphing utility to check the result numerically

Answer to Problem 19E

True

Explanation of Solution

Given:

Equation: $\frac{1}{\tan x}+\frac{1}{\cot x}=\tan x+\cot x$

Formula used:

  $\cot x=\frac{1}{\tan x}$

Calculation:

1. Algebraic method

Consider Left Hand side of the given equation,

  $\Rightarrow \frac{1}{\tan x}+\frac{1}{\cot x}$

Rewriting cot in terms of tan using identity, $\cot x=\frac{1}{\tan x}$ ,

  $\begin{array}{l}\Rightarrow \frac{1}{\tan x}+\frac{1}{\frac{1}{\tan x}}\\ \Rightarrow \frac{1}{\tan x}+\tan x\mathrm{......}(i)\end{array}$

Now, consider Right Hand Side of the given equation,

  $\Rightarrow \tan x+\cot x$

Rewriting cot in terms of tan using identity, $\cot x=\frac{1}{\tan x}$ ,

  $\begin{array}{l}\Rightarrow \tan x+\frac{1}{\tan x}\\ \Rightarrow \frac{1}{\tan x}+\tan x\mathrm{......}(ii)\end{array}$

It is clear that, equation (i) and equation (ii) are equal.

Therefore, Left Hand Side is equal to Right Hand Side.

Hence, $\frac{1}{\tan x}+\frac{1}{\cot x}=\tan x+\cot x$ .

2. Numerical method

Let, $f(x)=\frac{1}{\tan x}+\frac{1}{\cot x}\text{and}g(x)=\tan x+\cot x$ .

Now, using the table feature of graphing utility,

  

Interpretation:

From the above table, it is clear that, for any value of x, both functions were resulting same value.

So, both functions are equal to each other.

Conclusion:

Therefore, $\frac{1}{\tan x}+\frac{1}{\cot x}=\tan x+\cot x$ .