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2 cot x plus 1 equals -1?
2 cot(x) + 1 = -1 2 cot(x) = -2 cot(x) = -1 cos(x)/sin(x) = -1 cos(x) = - sin(x) x = 135°, 315°, 495°, ... another one every 180 degrees
What is the simplified expression of (1-cot(x))squaredcot(x)?
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)).
How do you prove that (1 plus cotx)2-2cotx 1(1-cos)(1 plus cos)?
Manipulate normally, noting:cot x = cos x / sin xcos² x + sin² x = 1 → sin²x = 1 - cos² xa² - b² = (a + b)(a - b)1 = 1²ab = baa/(bc) = a/b/c(1 + cot x)² - 2 cot x = 1² + 2 cot x + cot² x - 2 cot x= 1 + cot² x= 1 + (cos x / sin x)²= 1 + cos² x / sin² x= 1 + cos² x / (1 - cos² x)= ((1 - cos² x) + cos² x)/(1 - cos² x)= 1/(1² - cos² x)= 1/((1 + cos x)(1 - cos x))= 1/(1 - cos x)/(1 + cos x)QED.
What is the second derivative of ln(tan(x))?
f'(x) = 1/tan(x) * sec^2(x) where * means multiply and ^ means to the power of. = cot(x) * sec^2(x) f''(x) = f'(cot(x)*sec^2(x) + cot(x)*f'[sec^2(x)] = -csc^2(x)*sec^2(x) + cot(x)*2tan(x)sec^2(x) = sec^2(x) [cot(x)-csc^2(x)] +2tan(x)cot(x) = sec^2(x) [cot(x)-csc^2(x)] +2
Cos2x equals 1 minus tan squared x divide by1 plus tan squared x?
The Answer is 1 coz, 1-Tan squarex = Cot square X. So cot square x divided cot square x is equal to 1
2 cot x plus 1 equals -1?
2 cot(x) + 1 = -1 2 cot(x) = -2 cot(x) = -1 cos(x)/sin(x) = -1 cos(x) = - sin(x) x = 135°, 315°, 495°, ... another one every 180 degrees
What is the simplified expression of (1-cot(x))squaredcot(x)?
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)).
How do you prove that (1 plus cotx)2-2cotx 1(1-cos)(1 plus cos)?
Manipulate normally, noting:cot x = cos x / sin xcos² x + sin² x = 1 → sin²x = 1 - cos² xa² - b² = (a + b)(a - b)1 = 1²ab = baa/(bc) = a/b/c(1 + cot x)² - 2 cot x = 1² + 2 cot x + cot² x - 2 cot x= 1 + cot² x= 1 + (cos x / sin x)²= 1 + cos² x / sin² x= 1 + cos² x / (1 - cos² x)= ((1 - cos² x) + cos² x)/(1 - cos² x)= 1/(1² - cos² x)= 1/((1 + cos x)(1 - cos x))= 1/(1 - cos x)/(1 + cos x)QED.
Cotangent squared plus one equals cosecant squared?
yes 1 + cot x^2 = csc x^2
What is cot x?
Cot x is 1/tan x or cos x / sin x or +- sqrt cosec^2 x -1
Prove identity cos 2xcot2 x-1cot2 x1?
the questions is 2x=(cot^2 x-1)/(cot^2 x+1)
What is the second derivative of ln(tan(x))?
f'(x) = 1/tan(x) * sec^2(x) where * means multiply and ^ means to the power of. = cot(x) * sec^2(x) f''(x) = f'(cot(x)*sec^2(x) + cot(x)*f'[sec^2(x)] = -csc^2(x)*sec^2(x) + cot(x)*2tan(x)sec^2(x) = sec^2(x) [cot(x)-csc^2(x)] +2tan(x)cot(x) = sec^2(x) [cot(x)-csc^2(x)] +2
Cos2x equals 1 minus tan squared x divide by1 plus tan squared x?
The Answer is 1 coz, 1-Tan squarex = Cot square X. So cot square x divided cot square x is equal to 1
What is the derivative of x equals cot2 divided by t?
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
Csc squared divided by cot equals csc x sec. can someone make them equal?
cot(x)=1/tan(x)=1/(sin(x)/cos(x))=cos(x)/sin(x) csc(x)=1/sin(x) sec(x)=1/cos(x) Therefore, (csc(x))2/cot(x)=(1/(sin(x))2)/cot(x)=(1/(sin(x))2)/(cos(x)/sin(x))=(1/(sin(x))2)(sin(x)/cos(x))=(1/sin(x))*(1/cos(x))=csc(x)*sec(x)
What is the answer to cot squared x - tan squared x equals 0?
cot2x-tan2x=(cot x -tan x)(cot x + tan x) =0 so either cot x - tan x = 0 or cot x + tan x =0 1) cot x = tan x => 1 / tan x = tan x => tan2x = 1 => tan x = 1 ou tan x = -1 x = pi/4 or x = -pi /4 2) cot x + tan x =0 => 1 / tan x = -tan x => tan2x = -1 if you know about complex number then infinity is the solution to this equation, if not there's no solution in real numbers.
What is the derivative of x-4cscx 2cotx?
To find the derivative of the function ( f(x) = x - 4 \csc(x) \cdot 2 \cot(x) ), we first differentiate each term separately. The derivative of ( x ) is ( 1 ). For the second term, we apply the product rule: the derivative of ( -4 \csc(x) \cdot 2 \cot(x) ) involves differentiating ( -4 \csc(x) ) and ( 2 \cot(x) ), resulting in ( -4(2(-\csc(x)\cot^2(x) - \csc^2(x))) ). Thus, the complete derivative is ( f'(x) = 1 - 4 \left( 2(-\csc(x)\cot^2(x) - \csc^2(x)) \right) ).
