## GOODMAN 3 years ago Verify that this is an identity. sec(beta)+csc(beta) All over: 1+tan(beta) is equal to=csc(beta)

1. GOODMAN

|dw:1337050670476:dw|

2. lgbasallote

hmm maybe rewrite tan B as $$\frac{\sin B}{\cos B}$$?

3. Callisto

$\frac{sec\beta + csc\beta}{1+tan\beta} = \frac{\frac{sin\beta+cos\beta}{sin\beta cos\beta}}{\frac{sin\beta + cos\beta}{cos\beta}} = \frac{1}{sin\beta}=csc\beta$

4. GOODMAN

lol..thats soo small.

5. Callisto

How can I make it bigger?

6. lgbasallote

$\frac{\frac{1}{\cos B} + \frac{1}{\sin B}}{\frac{\cos B + \sin B}{\cos B}}$ $\frac{\frac{\sin B + \cos B}{\sin B \cos B}}{\frac{\cos B + \sin B}{\cos B}}$ $\frac{\frac{1}{\sin B \cos B}}{\frac{1}{\cos B}}$

7. lgbasallote

ugh @Callisto said it :P

8. GOODMAN

No..wait..im still a little confused, how did you guys split it?

9. lgbasallote

$\large \frac{1}{\sin B} = \csc B$

10. lgbasallote

split what?

11. GOODMAN

Yes..i know that, i know the basic identities.

12. GOODMAN

I mean, like the second and third step you did.

13. lgbasallote

$\large \sec B + \csc B = \frac{1}{\cos B} + \frac{1}{\sin B}$ LCD $\frac{\sin B + \cos B}{\sin B \cos B}$

14. Callisto

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

15. lgbasallote

$1 + \tan B = \frac{\cos B}{\cos B} + \frac{\sin B}{\cos B}$

16. lgbasallote

$\frac{\sin B + \cos B}{\cos B}$

17. Callisto

$1+tan\beta = 1+ \frac{sin\beta}{cos\beta} = \frac{cos\beta}{cos\beta} +\frac{sin\beta}{cos\beta} = \frac{sin\beta+cos \beta}{cos\beta}$

18. lgbasallote

the $$\sin B + \cos B$$ cancels

19. GOODMAN

Oh..okayy...basically i had to lay out the fractions? Arghh this stuff is weird.

20. lgbasallote

$\frac{\frac{1}{\sin B \cos B}}{\frac{1}{\cos B}}$$\frac{\cos B}{\sin B \cos B}$

21. Callisto

When you make it like that, it's easier to simplify~ Maths is weird in nature... I think :S

22. GOODMAN

Okay, I think i got it from here, thanks :D