# I need some serious help: trig expressions, OY!

1. Aug 8, 2008

### Wholewheat458

Here is my problem:
(sec 2 x csc x) / (sec 2 x + csc 2 x)
-first i replaced the denominator with its identities.. and i got : 2+tan^2 x +cos^2 x
but the numerator.. can you just .. i really think i'm doing this
wrong.. ugh, and i cannot stand it!!
Honestly, i would be truly gratful for any help ^^
:shy:

2. Aug 8, 2008

### snipez90

Um, is there an actual identity this expression is a part of? That's not really a problem, it's just an expression. Are you being asked to transform it into a nicer form? There are many, many possibilities for different expressions that are equivalent to the one you gave.

3. Aug 8, 2008

### Wholewheat458

^^ oh, i'm actually suposed to simplify the expression in the most complete form

4. Aug 8, 2008

### snipez90

Ok and to double check your original expression is [sec(2x)csc(x)] / [sec(2x) + csc(2x)] right?

5. Aug 8, 2008

### Wholewheat458

oh,.. no, i forgot the carrots sorry, the sec,sec, and csc are all squared
(sec^2 x csc x) / (sec^2 x + csc^2 x)

6. Aug 8, 2008

### snipez90

Ok, first, try writing everything in terms of sin and cos (often a good technique). The key lies in simplifying the denominator since you can't do much with the numerator after writing it in terms of sin and cos.

Here's a hint for the denominator. Since we're dealing with sums of reciprocals, use the fact that 1/a + 1/b = (a+b)/ab.

Last edited: Aug 8, 2008
7. Aug 8, 2008

### Wholewheat458

^^ WOOT!! Hahahhaha!! the answer is sin x!!!!
thank you so much!!

8. Aug 8, 2008

### snipez90

No problem! Also, this is more trig and less calculus and you're likely to get more help in the precalculus section of the homework area anyways. That way you'll get your problem answered faster without confusing people :).