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...
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.
PS: Here was my original post...