To prove the identity:

$\tanh^2(x) + \sech^2(x) = 1,$

we can start from the definitions of the hyperbolic tangent ($\tanh(x)$) and hyperbolic secant ($\sech(x)$) functions:

1. Definition of hyperbolic functions:

   • $\tanh(x) = \frac{\sinh(x)}{\cosh(x)}$,
   • $\sech(x) = \frac{1}{\cosh(x)}$, where $\sinh(x) = \frac{e^x - e^{-x}}{2}$ and $\cosh(x) = \frac{e^x + e^{-x}}{2}$.

2. Substituting definitions: Substitute $\tanh(x)$ and $\sech(x)$ into $\tanh^2(x) + \sech^2(x)$: $\tanh^2(x) + \sech^2(x) = \left(\frac{\sinh(x)}{\cosh(x)}\right)^2 + \left(\frac{1}{\cosh(x)}\right)^2.$

3. Simplifying: Rewrite using a common denominator ($\cosh^2(x)$): $\tanh^2(x) + \sech^2(x) = \frac{\sinh^2(x)}{\cosh^2(x)} + \frac{1}{\cosh^2(x)}.$

   Combine terms: $\tanh^2(x) + \sech^2(x) = \frac{\sinh^2(x) + 1}{\cosh^2(x)}.$

4. Hyperbolic identity: Recall the fundamental hyperbolic identity: $\cosh^2(x) - \sinh^2(x) = 1.$

   Rearrange it: $\cosh^2(x) = \sinh^2(x) + 1.$

   Substitute $\cosh^2(x)$ back into the expression: $\tanh^2(x) + \sech^2(x) = \frac{\cosh^2(x)}{\cosh^2(x)} = 1.$

5. Conclusion: Therefore, the identity $\tanh^2(x) + \sech^2(x) = 1$ is proven.

Would you like further clarification on any step? Here are 5 related questions to deepen your understanding:

1. How are hyperbolic functions related to their trigonometric counterparts?
2. Can you derive the fundamental hyperbolic identity $\cosh^2(x) - \sinh^2(x) = 1$ from exponential definitions?
3. What are other identities involving $\tanh(x)$ and $\sech(x)$?
4. How can hyperbolic identities be used in solving differential equations?
5. Can you explore the graphs of $\tanh(x)$ and $\sech(x)$ to understand their behavior visually?

Tip: Memorize the key hyperbolic identities—they often simplify otherwise complex problems!