Question

If $x=\frac{a^n-a^{-n}}{a^n+a^{-n}}$ then at $n=1$, $x=\frac{a^2-1}{a^2+1}$

If above statement is true enter $1$ else enter $0$.

Answer 1

Given, $x=\frac{a^n-a^{-n}}{a^n+a^{-n}}$

$n=1$

put the value on $n$ in given equation,

So,

$x=\frac{a^1-a^{-1}}{a^1+a^{-1}}$

$x=\frac{a-\frac{1}{a}}{a+\frac{1}{a}}$

$x=\frac{\frac{a^2-1}{a}}{\frac{a^2+1}{a}}$

$x=\frac{a^2-1}{a^2+1}$

The given statement is true.