in second and fourth... for angles 135 and 315 degrees
To simplify the expression ((1 - \cot(x))^2 \cot(x)), we start by expanding ((1 - \cot(x))^2) to get (1 - 2\cot(x) + \cot^2(x)). Then, we multiply this by (\cot(x)): [ (1 - 2\cot(x) + \cot^2(x)) \cot(x) = \cot(x) - 2\cot^2(x) + \cot^3(x). ] Thus, the simplified expression is (\cot(x) - 2\cot^2(x) + \cot^3(x)).
csc^2x+cot^2x=1
Quadrant 3 Quadrant 4 Quadrant 2 Quadrant 1
-1
Assuming you want dx/dt and that the equation is x = cot(2) / t (i.e. cot(2) is a constant) we can use the power rule. First, we rewrite it: cot(2)/t = cot(2) * t-1 thus, by the power rule: dx/dt = (-1) cot(2) * t-1 -1 = - cot(2) * t-2= = -cot(2)/t2
It is -sqrt(1 + cot^2 theta)
Cotan(theta) is the reciprocal of the tan(theta). So, cot(theta) = 1/2.
To simplify the expression ((1 - \cot(x))^2 \cot(x)), we start by expanding ((1 - \cot(x))^2) to get (1 - 2\cot(x) + \cot^2(x)). Then, we multiply this by (\cot(x)): [ (1 - 2\cot(x) + \cot^2(x)) \cot(x) = \cot(x) - 2\cot^2(x) + \cot^3(x). ] Thus, the simplified expression is (\cot(x) - 2\cot^2(x) + \cot^3(x)).
1 quadrant = 1 quadrant. Or what is the question?
To find the quadrant and sign of the cotangent function for -495 degrees, first, convert it to a positive angle by adding 360 degrees until the angle is within the standard range. -495 + 720 = 225 degrees. The angle 225 degrees is in the third quadrant, where both sine and cosine are negative, making cotangent (which is the ratio of cosine to sine) positive. Thus, cot(-495 degrees) is positive and located in the third quadrant.
Quadrant 1: (1,5) Quadrant 2: (-2,3) Quadrant 3: (-3,-3) Quadrant 4:(4,-1)
csc^2x+cot^2x=1
Quadrant 3 Quadrant 4 Quadrant 2 Quadrant 1
5
quadrant 1
-1
Assuming you want dx/dt and that the equation is x = cot(2) / t (i.e. cot(2) is a constant) we can use the power rule. First, we rewrite it: cot(2)/t = cot(2) * t-1 thus, by the power rule: dx/dt = (-1) cot(2) * t-1 -1 = - cot(2) * t-2= = -cot(2)/t2