Answer
$\frac{1}{t^2+9}$
Work Step by Step
$\frac{t-3}{t^2+9}\times\frac{t+3}{t^2-9}$
Factorise the fraction $\frac{t-3}{t^2-9}$:
$=\frac{t-3}{t^2+9}\times\frac{t+3}{(t-3)(t+3)}$
Multiply the fractions:
$=\frac{(t-3)(t+3)}{(t^2+9)(t-3)(t+3)}$
Simplify the fraction:
$=\frac{1}{t^2+9}$
