Let's solve the equation step by step:

Given equation:

\[
\frac{1}{\cos x + 1} + \frac{1}{\cos x - 1} = -2 \csc x \cot x
\]

### Step 1: Combine the left-hand side (LHS)

The LHS contains two fractions. To combine them, we need a common denominator:

\[
\frac{1}{\cos x + 1} + \frac{1}{\cos x - 1} = \frac{(\cos x - 1) + (\cos x + 1)}{(\cos x + 1)(\cos x - 1)}
\]

Simplify the numerator:

\[
(\cos x - 1) + (\cos x + 1) = 2\cos x
\]

Now the LHS becomes:

\[
\frac{2\cos x}{(\cos x + 1)(\cos x - 1)}
\]

Recall the difference of squares formula:

\[
(\cos x + 1)(\cos x - 1) = \cos^2 x - 1
\]

Thus, the LHS becomes:

\[
\frac{2\cos x}{\cos^2 x - 1}
\]

### Step 2: Simplify the right-hand side (RHS)

The RHS is given as:

\[
-2 \csc x \cot x
\]

We know the identities for \(\csc x\) and \(\cot x\):

\[
\csc x = \frac{1}{\sin x}, \quad \cot x = \frac{\cos x}{\sin x}
\]

So, the RHS becomes:

\[
-2 \cdot \frac{1}{\sin x} \cdot \frac{\cos x}{\sin x} = -2 \frac{\cos x}{\sin^2 x}
\]

### Step 3: Set both sides equal

Now we have:

\[
\frac{2\cos x}{\cos^2 x - 1} = -2 \frac{\cos x}{\sin^2 x}
\]

### Step 4: Simplify the equation

First, cancel \(2\cos x\) from both sides (assuming \(\cos x \neq 0\)):

\[
\frac{1}{\cos^2 x - 1} = -\frac{1}{\sin^2 x}
\]

We know that \(\cos^2 x + \sin^2 x = 1\), which implies:

\[
\cos^2 x - 1 = -\sin^2 x
\]

Substitute this into the LHS:

\[
\frac{1}{-\sin^2 x} = -\frac{1}{\sin^2 x}
\]

Thus, both sides are equal, confirming that the equation is **true**.

### Conclusion:

The given equation is an identity and holds true for all values of \(x\) (except where the trigonometric functions are undefined, like when \(\sin x = 0\)).

Would you like more details or clarifications?

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Here are 5 related questions for further exploration:

1. What is the derivation of the cotangent and cosecant functions in terms of sine and cosine?
2. How does the identity \(\cos^2 x + \sin^2 x = 1\) help in simplifying trigonometric equations?
3. What are the conditions where trigonometric functions like \(\csc x\) and \(\cot x\) are undefined?
4. How can you use the difference of squares formula in trigonometry?
5. Can this identity be verified using specific values of \(x\) (e.g., \(x = \frac{\pi}{4}\))?

**Tip**: Always remember that certain trigonometric functions like \(\csc x\) and \(\cot x\) have undefined points where \(\sin x = 0\).

Explore the detailed step-by-step solution to the trigonometric identity 1/(cosx+1) + 1/(cosx-1) = -2cscxcotx. This problem covers key trigonometric concepts such as identities, Pythagorean identity, and simplifying trigonometric fractions. Suitable for students in Grades 10-12.

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