## anonymous one year ago helpppppppp please.

1. anonymous

2. anonymous

@Hero

3. DanJS

What do you want to do with that?

4. anonymous

Simplify the trigonometric expression. Show your work.

5. DanJS

if you want to combine the two fractions, need to multiply by a common denominator

6. anonymous

okay... idk how to do that?

7. DanJS

$[\frac{ 1 }{ 1+\sin \theta }+\frac{ 1 }{ 1-\sin \theta }]*\frac{ (1+\sin \theta)(1-\sin \theta) }{ (1+\sin \theta)(1-\sin \theta) }$

8. DanJS

Just multiplying the whole thing by 1... a ratio of something over itself

9. anonymous

okay.. what does sin o stand for?

10. anonymous

is it kinda like pi=3.14? does it stand for a number?

11. anonymous

can u do it step by step and explain..

12. jdoe0001

one sec

13. jdoe0001

$$\bf \cfrac{1}{1+sin(\theta)}+\cfrac{1}{1-sin(\theta)}\impliedby LCD\to [1+sin(\theta)][1-sin(\theta)] \\ \quad \\ \cfrac{(1-sin(\theta))+(1+sin(\theta))}{[1+sin(\theta)][1-sin(\theta)]}\implies \cfrac{1\cancel{-sin(\theta)}+1\cancel{+sin(\theta)}}{[1+sin(\theta)][1-sin(\theta)]} \\ \quad \\ ----------------------------------\\ \textit{difference of squares} \\ \quad \\ (a-b)(a+b) = a^2-b^2\qquad \qquad a^2-b^2 = (a-b)(a+b)\qquad thus\\ ---------------------------------- \\ \quad \\ \cfrac{1\cancel{-sin(\theta)}+1\cancel{+sin(\theta)}}{{\color{brown}{ [1^2-sin^2(\theta)] }}}\implies \cfrac{2}{{\color{brown}{ cos^2(\theta)}}} \\ \quad \\ 2\cdot \cfrac{1}{cos^2(\theta)}\implies 2sec^2(\theta)$$

14. jdoe0001

the denominator is combined, since it's just a difference of squares and keep in mind that $$\bf sin^2(\theta)+cos^2(\theta)=1\implies sin^2(\theta)={\color{brown}{ 1-cos^2(\theta)}}$$

15. jdoe0001

woops, darn that came out off lemme redo that

16. jdoe0001

$$\bf sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)={\color{brown}{ 1-sin^2(\theta)}}$$ rather