prove using osbornes rule\n1. $coth^{2}x - 1=cosech^{2}x$

Explanation:

Step1: Recall hyperbolic function definitions

We know that $\coth x=\frac{\cosh x}{\sinh x}$ and $\text{cosech }x = \frac{1}{\sinh x}$.

Step2: Substitute into left - hand side

Substitute $\coth x=\frac{\cosh x}{\sinh x}$ into $\coth^{2}x - 1$. We get $\left(\frac{\cosh x}{\sinh x}\right)^{2}-1=\frac{\cosh^{2}x}{\sinh^{2}x}-1$.

Step3: Get a common denominator

$\frac{\cosh^{2}x}{\sinh^{2}x}-1=\frac{\cosh^{2}x-\sinh^{2}x}{\sinh^{2}x}$.

Step4: Use the identity $\cosh^{2}x-\sinh^{2}x = 1$

Since $\cosh^{2}x-\sinh^{2}x = 1$, then $\frac{\cosh^{2}x-\sinh^{2}x}{\sinh^{2}x}=\frac{1}{\sinh^{2}x}$.

Step5: Recall the definition of $\text{cosech }x$

Since $\text{cosech }x=\frac{1}{\sinh x}$, then $\frac{1}{\sinh^{2}x}=\text{cosech}^{2}x$.

Answer:

We have shown that $\coth^{2}x - 1=\text{cosech}^{2}x$ by using the definitions of hyperbolic functions and the identity $\cosh^{2}x-\sinh^{2}x = 1$.