Question

$\coth^2 x - \tanh^2 x =$

• A. $4\text{cosech}2x \tanh 2x$
• B. $4\text{sech}2x \coth 2x$
• C. $4\text{sech}2x \tanh 2x$
• D. $4\cosh^2 x (\text{cosech}2x)^2$

Hint:

Master the fundamental hyperbolic identities: $\cosh^2 x - \sinh^2 x = 1$, $1 - \tanh^2 x = \text{sech}^2 x$, $\coth^2 x - 1 = \text{cosech}^2 x$. Also know the double angle formulas: $\sinh(2x)=2\sinh x\cosh x$ and $\cosh(2x)=\cosh^2 x+\sinh^2 x$.

Answer 1

The Correct Option is D

We need to simplify the expression $\coth^2 x - \tanh^2 x$.
First, express $\coth x$ and $\tanh x$ in terms of $\cosh x$ and $\sinh x$.
$\coth x = \frac{\cosh x}{\sinh x}$ and $\tanh x = \frac{\sinh x}{\cosh x}$.
The expression becomes:
$(\frac{\cosh x}{\sinh x})^2 - (\frac{\sinh x}{\cosh x})^2 = \frac{\cosh^2 x}{\sinh^2 x} - \frac{\sinh^2 x}{\cosh^2 x}$.
Combine the fractions by finding a common denominator:
$\frac{\cosh^4 x - \sinh^4 x}{\sinh^2 x \cosh^2 x}$.
The numerator is a difference of squares: $\cosh^4 x - \sinh^4 x = (\cosh^2 x - \sinh^2 x)(\cosh^2 x + \sinh^2 x)$.
Using the fundamental hyperbolic identities:
$\cosh^2 x - \sinh^2 x = 1$.
$\cosh^2 x + \sinh^2 x = \cosh(2x)$.
So, the numerator simplifies to $1 \cdot \cosh(2x) = \cosh(2x)$.
The denominator can be related to $\sinh(2x)$. We know $\sinh(2x) = 2 \sinh x \cosh x$.
Therefore, $(\sinh x \cosh x)^2 = (\frac{\sinh(2x)}{2})^2 = \frac{\sinh^2(2x)}{4}$.
Substituting these back into the expression:
$\frac{\cosh(2x)}{\frac{\sinh^2(2x)}{4}} = \frac{4\cosh(2x)}{\sinh^2(2x)}$.
This does not match any of the options directly. Let's re-evaluate the options. The provided key is (D). Let's simplify (D).
$4\cosh^2 x (\text{cosech}2x)^2 = 4\cosh^2 x \left(\frac{1}{\sinh 2x}\right)^2 = \frac{4\cosh^2 x}{\sinh^2 2x} = \frac{4\cosh^2 x}{(2\sinh x \cosh x)^2} = \frac{4\cosh^2 x}{4\sinh^2 x \cosh^2 x} = \frac{1}{\sinh^2 x} = \coth^2 x$. This is not correct. There is an error.
Let's re-simplify the original expression in another way.
$\coth^2 x - \tanh^2 x = (1+\text{cosech}^2 x) - (1-\text{sech}^2 x) = \text{cosech}^2 x + \text{sech}^2 x$.
$= \frac{1}{\sinh^2 x} + \frac{1}{\cosh^2 x} = \frac{\cosh^2 x + \sinh^2 x}{\sinh^2 x \cosh^2 x} = \frac{\cosh(2x)}{(\frac{1}{2}\sinh(2x))^2} = \frac{4\cosh(2x)}{\sinh^2(2x)}$.
This result is consistent. All options seem to be incorrect, or there is a typo in the question or options. Let's re-examine option (D) from the image: $4\cosh2x (\text{cosech}2x)^2$. OCR was $4\cosh^2 x$. The image has $\cosh 2x$.
Let's simplify $4\cosh 2x (\text{cosech}2x)^2$. $4\cosh 2x (\frac{1}{\sinh 2x})^2 = \frac{4\cosh 2x}{\sinh^2 2x}$.
This exactly matches my simplification. So the OCR had a typo, but the keyed answer (with the corrected term from the image) is likely correct.