Question

Prove that $tanh ( x - y) = \frac{tanh\ x\ -\ tanh\ y}{1\ -\ tanh\ x\ .\ tanh\ y}$.

Answer 1

Hi.

The given function is a Hyperbolic Trigonometric Function.

I have used the formulas of Hyperbolic Trigonometric Functions to solve the question.

I have proved it using simple laws of mathematics.

I have provided its proof in detail in the ATTACHMENT.

Kindly see the attachment for detailed answer.

I hope it will help You.

Answer 2

HELLO DEAR,

we know:- tanha = $\frac{e^a - e^{-a}}{e^a + e^{-a}}$

now, $\frac{tanh\ x\ -\ tanh\ y}{1\ -\ tanh\ x\ .\ tanh\ y}$

$\bold{\implies \frac{(e^x - e^{-x})/(e^x + e^{-x}) - (e^y - e^{-y})/(e^y + e^{-y})}{1 - (e^x - e^{-x})/(e^x + e^{-x}) \times (e^y - e^{-y})/(e^y + e^{-y})}}$

$\bold{\implies \frac{e^{x + y} + e^{x - y} - e^{-x + y} - e^{-x - y} - e^{x + y} + e^{x - y} - e^{-x + y} + e^{-x - y}}{e^{x + y} + e^{x - y} + e^{y - x} + e^{-x - y} - e^{x + y} + e^{x - y} + e^{y - x} - e^{-x - y}}}$

$\implies \bold{\frac{\cancel{e^{x + y}} + e^{x - y} - e^{-x + y} - \cancel{e^{-x - y}} - \cancel{e^{x + y}} + e^{x - y} - e^{y - x} + \cancel{e^{-x - y}}}{\cancel{e^{x + y}} + e^{x - y} + e^{y - x} + \cancel{e^{-x - y}} - \cancel{e^{x + y}} + e^{x - y} + e^{y - x} - \cancel{e^{-x - y}}}}$

$\implies \bold{\frac{2\{e^{x - y} - e^{-x + y}\}}{2\{e^{x - y} + e^{-x + y}\}}}$

$\implies \bold{\frac{e^{x - y} - e^{-x + y}}{e^{x - y} + e^{-x + y}} = tanh(x - y)}$


I HOPE IT'S HELP YOU DEAR,
THANKS