Chapter 5.2, Problem 23E

To determine

To verify the identity $\frac{1}{\cos x+1}+\frac{1}{\cos x-1}=-2\csc x\cot x$ algebraically and to use table feature of graphing utility to check the result numerically

Answer to Problem 23E

True

Explanation of Solution

Given:

Equation: $\frac{1}{\cos x+1}+\frac{1}{\cos x-1}=-2\csc x\cot x$

Formula used:

  $\begin{array}{l}\frac{\cos x}{sinx}=\cot x\\ \frac{1}{\sin x}=\csc x\end{array}$

Calculation:

1. Algebraic method

Consider Left Hand side of the given equation,

  $\begin{array}{l}\Rightarrow \frac{1}{\cos x+1}+\frac{1}{\cos x-1}\\ \Rightarrow \frac{\left(\cos x-1\right)+\left(\cos x+1\right)}{\left(\cos x+1\right)\left(\cos x-1\right)}[\because \text{Taking Least Common Factor}]\\ \Rightarrow \frac{2\cos x}{\left(\cos x+1\right)\left(\cos x-1\right)}\\ \Rightarrow \frac{2\cos x}{{\cos }^{2}x-1}\\ \Rightarrow \frac{2\cos x}{1-si{n}^{2}x-1}[\because {\cos }^{2}x=1-{\sin }^{2}x]\\ \Rightarrow \frac{2\cos x}{-si{n}^{2}x}\end{array}$

  $\begin{array}{l}\Rightarrow -2\left(\frac{1}{\sin x}\right)\left(\frac{\cos x}{sinx}\right)[\because \frac{\cos x}{sinx}=\cot x,\frac{1}{\sin x}=\csc x]\\ \Rightarrow -2\csc x\cot x\mathrm{......}(i)\end{array}$

Now, consider Right Hand Side of the given equation,

  $\Rightarrow -2\csc x\cot x\mathrm{......}(ii)$

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

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

Hence, $\frac{1}{\cos x+1}+\frac{1}{\cos x-1}=-2\csc x\cot x$ .

2. Numerical method

Let, $f(x)=\frac{1}{\cos x+1}+\frac{1}{\cos x-1}\text{and}g(x)-2\csc x\cot x$ .

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, $\frac{1}{\cos x+1}+\frac{1}{\cos x-1}=-2\csc x\cot x$ .