Eqn 11e Help
Given that ex/a = u + √(1+u2), we now invert both sides, which remain equal.
LHS: 1 / ex/a = e–x/a
| RHS: 1 / {u + √(1+u2)} | = {u – √(1+u2)} / [{u – √(1+u2)} • {u + √(1+u2)}] | |
| = {u – √(1+u2)} / [u2 – (1+u2)] | ||
| = {u – √(1+u2)} / [–1] | ||
| = – u + √(1+u2) | ||
| So | e–x/a | = – u + √(1+u2) |
The technique used to clear the surdic denominator from the inverted RHS is frequently used in surdical operations, and is an ingenious variation on the difference between two squares: b2 – c = (b + √c) • (b – √c)