Solve the given question, using following laws of indices,

Laws of Indices,

1-: a^m × aⁿ = a^{{m + n}}

2-: a^m ÷ aⁿ = a^{{m - n}}

3-: [(a^m)ⁿ] = a^mn

4-: (a)^1/m = $(\sqrt[m]{a})$

5-: (a)^-m = 1/a^m

6-: (a)^(m/n) = $(\sqrt[n]{{{a^m}}})$

7-: a⁰ = 1

⇒ 1 - {1 + (a² -1)^-1}^-1

$(\begin{array}{l} \Rightarrow 1 - {\left\{ {1 + \frac{1}{{{a^{2\;}} - 1}}} \right\}^{ - 1}}\\ \Rightarrow 1 - {\left\{ {\frac{{{a^2}}}{{{a^2} - 1}}} \right\}^{ - 1}}\\ \Rightarrow 1 - \left\{ {\frac{{{a^2}}}{{{a^2} - 1}}} \right\}^{ - 1}\\ \Rightarrow \frac{1}{{{a^2}}} \end{array})$