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Express csc theta in terms of cot theta theta is in quadrant 3?
It is -sqrt(1 + cot^2 theta)
How do you solve csc x sin x equals cos x cot x plus?
Suppose csc(x)*sin(x) = cos(x)*cot(x) + y then, ince csc(x) = 1/sin(x), and cot(x) = cos(x)/sin(x), 1 = cos(x)*cos(x)/sin(x) + y so y = 1 - cos2(x)/sin(x) = 1 - [1 - sin2(x)]/sin(x) = [sin2(x) + sin(x) - 1]/sin(x)
How do you simplify csc theta cot theta cos theta?
cosec(q)*cot(q)*cos(q) = 1/sin(q)*cot(q)*cos(q) = cot2(q)
How do you find the derivative of - csc x - sin x?
d/dx (-cscx-sinx)=cscxcotx-cosx
Write the expression in terms of sine and cosine and simplify so that no quotients appear in the final expression. cscx(sinx plus cosx)?
csc(x)*{sin(x) + cos(x)} = csc(x)*sin(x) + csc(x)*cos(x) =1/sin*(x)*sin(x) + 1/sin(x)*cos(x) = 1 + cot(x)
What is the derivative of csc x?
The derivative of csc(x) is -cot(x)csc(x).
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) ).
What is the derivative of 3tanx-4cscx?
7
What is the anti-derivative of co secant x?
According to Wolfram Alpha, input:integral csc x it is -log[cot(x) + csc(x)] + constant You can verify this by taking the derivative of the purported integral.
What is the anti derivative of cscxcotx?
∫cscxcotx*dx∫csc(u)cot(u)*du= -csc(u)+C, where C is the constant of integrationbecause d/dx(csc(u))=-[csc(u)cot(u)],so d/dx(-csc(u))=csc(u)cot(u).∫cscxcotx*dxLet:u=xdu/dx=1du=dx∫cscucotu*du= -csc(u)+CPlug in x for u.∫cscxcotx*dx= -csc(x)+C
Csc divided by cot squared equals tan multiplied by sec?
Yes.
What is the derivitive of cenat?
The derivative of the function ( \csc(x) ) (cosecant) is given by ( -\csc(x) \cot(x) ). If you meant a different function by "cenat," please clarify, as "cenat" doesn't correspond to a standard mathematical term.
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
Can you simplify 1-cot x?
csc^2x+cot^2x=1
Cotangent squared plus one equals cosecant squared?
yes 1 + cot x^2 = csc x^2
How do you integrate the cscnx?
The integral for csc(u)dx is -ln|csc(u) + cot(u)| + C.
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)
