mtd

square root(a^2 -u^2) where a>0, let u=(a sinx) where -pi/2<x<pi/2 answer is (a cosx) and I don't know how to get there

sorry, another I can't figure out Show that (1-cot^2x)/(tan^2x-1)=cot^2x I started by factoring both as difference of squares. Would I be better served by writing in terms of sine and cosine? Such as: [1-(cos^2x/sin^2x)]/[(sin^2x/cos^2x)-1]=(cos^2x/sin^2x...

let me try to type it out again, my apologies. Original problem is--Prove the following: 1- cot^2 x ----------- = cot^2 x tan^2 x - 1 That is how the problem actually looks.

Take the two equation two unknown down to one unknown by either adding or subtracting the second equation from the first so that one of the uknowns is removed. 2x+y=5 + x-y=1 -------------- 3x=6 x=2 Then plug the newly found x value into one of the equati...

I believe the noble gas group or Group O. Hopefully someone can verify that. I apologize if that is wrong.