Re: Trig question pre calc!

Hey Gurp925.

Is it meant to be cos(theta) / [1 - sin(theta)] or cos(theta) - sin(theta)?

Re: Trig question pre calc!

$\displaystyle \frac{sin\theta+1-cos^2\theta}{cos\theta sin\theta}$

$\displaystyle =\frac{sin\theta+sin^2\theta}{cos\theta sin\theta}$

$\displaystyle =\frac{sin\theta(1+sin\theta)}{cos\theta sin\theta}$

$\displaystyle =\frac{1+sin\theta}{cos\theta}$

$\displaystyle =\frac{(1+sin\theta)(1-sin\theta)}{cos\theta(1-sin\theta)}$

$\displaystyle =\frac{1-sin^2\theta}{cos\theta(1-sin\theta)}$

$\displaystyle =\frac{cos^2\theta}{cos\theta(1-sin\theta)}$

...

Hope this helps :)

Re: Trig question pre calc!

Hey Acc100jt almost understood the question just wondering how did you multiply the denominator by (1-sin) ?thus multiplying the top? if i understand that then the question is solved. Thanks everyone for their help.

Re: Trig question pre calc!

Quote:

Originally Posted by

**Gurp925** Hey Acc100jt almost understood the question just wondering how did you multiply the denominator by (1-sin) ?thus multiplying the top? if i understand that then the question is solved. Thanks everyone for their help.

Multiplying top and bottom by the bottom's conjugate.