# Thread: Trig question pre calc!

1. ## Trig question pre calc!

Hi there, the question states

Prove algebraically: $\displaystyle \cos\theta/1-sin\theta$ = $\displaystyle sec\theta +(sec\theta)(csc\theta)-cot\theta$

Here are the steps i took to solving this i worked on the right side making it equal the left

1) change all sec/ csc/ cot to their respective forms ($\displaystyle \1/sin\theta)$ and so forth

and when i get to this point I'm lost $\displaystyle \frac{sin\theta +1 - cos\theta^2} {cos\theta sin\theta}$

After this point I'm lost please help, i apologize if i have asked too many questions but i am trying to study for my final.

Thanks!

2. ## Re: Trig question pre calc!

Hey Gurp925.

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

3. ## 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

4. ## 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.

5. ## Re: Trig question pre calc!

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.