 |
| ||
| |||
|
| Math Central | Quandaries & Queries |
|
I need help proving: 1-tan^3t/1-tant=sec^2t+tant |
Brian,
The pattern you need to recognize here is the difference of cubes expression
x3 - y3 = (x - y)(x2 + xy + y2)
In your problem I see this with x = 1 and y = tan(t).
Start with the left side of your expression and multiply the numerator and denominator by (1 + tan(t) + tan2(t)). Use the difference of cubes expression to simplify.
I hope this helps,
Harley
| ||
| ||
| ||
 |
 |
|
Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.