To solve the equation 

\[
\tanh^{-1}(\cos \theta) = \cosh^{-1}(\csc \theta),
\]

we can start by analyzing both sides of the equation separately.

### Left Side: \(\tanh^{-1}(\cos \theta)\)
The inverse hyperbolic tangent, \(\tanh^{-1}(x)\), is defined as:

\[
\tanh^{-1}(x) = \frac{1}{2} \ln\left(\frac{1 + x}{1 - x}\right)
\]

So,

\[
\tanh^{-1}(\cos \theta) = \frac{1}{2} \ln\left(\frac{1 + \cos \theta}{1 - \cos \theta}\right)
\]

### Right Side: \(\cosh^{-1}(\csc \theta)\)
The inverse hyperbolic cosine, \(\cosh^{-1}(x)\), is defined as:

\[
\cosh^{-1}(x) = \ln\left(x + \sqrt{x^2 - 1}\right)
\]

Thus,

\[
\cosh^{-1}(\csc \theta) = \ln\left(\csc \theta + \sqrt{\csc^2 \theta - 1}\right)
\]

### Equating Both Sides:
Now, equate the two expressions:

\[
\frac{1}{2} \ln\left(\frac{1 + \cos \theta}{1 - \cos \theta}\right) = \ln\left(\csc \theta + \sqrt{\csc^2 \theta - 1}\right)
\]

To eliminate the logarithms, exponentiate both sides:

\[
\left(\frac{1 + \cos \theta}{1 - \cos \theta}\right)^{\frac{1}{2}} = \csc \theta + \sqrt{\csc^2 \theta - 1}
\]

Square both sides:

\[
\frac{1 + \cos \theta}{1 - \cos \theta} = \left(\csc \theta + \sqrt{\csc^2 \theta - 1}\right)^2
\]

Expanding the right side:

\[
\frac{1 + \cos \theta}{1 - \cos \theta} = \csc^2 \theta + 2 \csc \theta \cdot \sqrt{\csc^2 \theta - 1} + (\csc^2 \theta - 1)
\]

Simplify further and try to isolate \(\theta\). However, solving the equation algebraically for \(\theta\) might require numerical methods or further trigonometric identities.

Would you like to go into further details about this solution or have any specific questions?

Here are 5 related questions to consider:
1. How is the hyperbolic tangent function related to trigonometric functions?
2. What are the properties of inverse hyperbolic functions?
3. Can the given equation be solved graphically for specific values of \(\theta\)?
4. How does the domain of the \(\tanh^{-1}(x)\) function affect the solution?
5. What are the implications of the equation for different ranges of \(\theta\)?

**Tip:** When dealing with inverse functions, always consider the domain and range to avoid extraneous solutions.

Explore the solution to the equation tanh^{-1}(cosθ) = cosh^{-1}(cscθ) through the analysis of hyperbolic and logarithmic functions. This detailed explanation covers key mathematical concepts and is suitable for students in Grades 11-12.

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