tan(3x) + 1 = sec(3x) Thanks, pretend 3x equals x so tanx + 1 = secx we know the law that 1 + tanx = secx so tanx + 1 becomes secx and... secx = secx sec(3x) = sec(3x) [just put 3x back in for x- you don't really have to change 3x
gahh i cant figure this one out either. you have you verify the identity only using one side of the equation: sec^2x*sin^2x+sec^2x=-1 this identity cannot possible be true. on the left side you are adding 2 squared quantities, how
Okay, I've been getting some of these, but I can't seem to verify this identity... any help? Here's the problem Sin(x+y) + Sin(x-y) = 2sinxcosy Okay, I've been working on the left side, and distribute, getting: Sinx + Siny + Sinx
it says to verify the following identity, working only on one side: cotx+tanx=cscx*secx Work the left side. cot x + tan x = cos x/sin x + sin x/cos x = (cos^2 x +sin^2x)/(sin x cos x) = 1/(sin x cos x) = 1/sin x * 1/cos x You're
Left side = right side check type of question. x - (x + y)/2 = (x - y)/2 What I did was times the x in the left side by 2. I got: =(2x-x+y)/2 which equals =(x+y)/2 But the left side is (x-y)/2. What am I doing wrong?