# Homework Help: Few questions on proving identities

1. Jan 16, 2008

### VanKwisH

1. The problem statement, all variables and given/known data

seca + csca
--------------- = seca*csca
sina + cosa

2. Relevant equations

no relevant equations

3. The attempt at a solution

How would i prove this ?? i know i break down the left side...
which is

1/cosa + 1/sina
---------------------
sina + cosa

2. Jan 16, 2008

### rocomath

Now break down the right side to sines and cosines. From there, cross multiply the denominator of your right side to the left, from there it will simplify to 1.

3. Jan 16, 2008

### VanKwisH

on the right side it's 1/cosa * 1/sina = 1 /cosasina right??

4. Jan 16, 2008

### rocomath

Yes, that's correct. Now cross multiply.

5. Jan 16, 2008

### VanKwisH

hmmm i did cross multiply and then i got ..............
((cosa*sina /cosa ) + ( cosa*sina /sina)) / sina+cosa for the left side
and then for the right side i got sina+cosa / cosa*sina for the right side ...... is that right??

6. Jan 16, 2008

### rocomath

Yes, that's correct. Now simplify and L=R.

7. Jan 16, 2008

### VanKwisH

alright for the left side i was able to get one but for the right side it doesn't seem right ....

sina + cosa / cosa*sina ......... that doesn't equal 1 does it.???

8. Jan 16, 2008

### rocomath

Sorry, all the writing confused me. The right side should only be sina+cosa, right? B/c you cross multiplied and so sinacosa disappears from the right.

9. Jan 16, 2008

### VanKwisH

OOOOOOOOOOOO alright ........ but that doesn't equal 1 though ........ only sin^2a + cos^2 a = 1

10. Jan 16, 2008

### VanKwisH

O wait but since i cross multiplied the denominator on the left side is gone as well right??

11. Jan 16, 2008

### HallsofIvy

I'm not sure what you mean by "cross multiply". To me, it means multiply both sides by the common denominator: here, that is (sin a cos a)(sin a+ cos a). You seem to have multiplied one side by (sin a+ cos a) and the other by (sin a cos a). Of course, that will not give you the same thing- you are multiplying the two sides by different things.

12. Jan 16, 2008

### VanKwisH

yahhh ......... i messed up but i fixed it now and i got it .........
but i think cross multiplying is different for example

sinx cosx
------ ---------
cosx sinx

from what i know ........ u would multiply the sinx on top by the one on the bottom ....
and the cosx by the cosx ...... therefore u would have sin^2x and cos^2x ........
that's how i interpret it......unlesss i am completely wrong ........