Trig help... (Prof's answer).

Hey guys... I logged on today only to find my other thread full of a bunch of arguing and it was also locked. I thought I'd post my Prof's reply for you guys to see what, exactly we are doing.

Here's what my prof. had to say...

Quote:

Your question is a good one. We must change the right side of the

eqution into a fraction. Yes, we need to add (or subtract)

1/sin^2(theta) - 1

So we need to have the same denominator to get

1/sin^2(theta) - sin^2(theta)/sin^2(theta)

Together we form the quotient [1 - sin^2(theta)] / sin^2(theta)

At this point you will have one quotient equaling another quotient, so

you can do the reciprocal to both sides of the equation; and then take

the square root on both sides.

-Tom

PS: Here was my original post...

http://www.mathhelpforum.com/math-help/21282-post1.html