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(tan x + cot x)/sec x . csc x
The key to solve this question is to turn tan x, cot x, sec x, csc x into the simpler form.
Remember that tan x = sin x / cos x, cot x = 1/tan x, sec x = 1/cos x, csc x = 1/sin x
The solution is:
[(sin x / cos x)+(cos x / sin x)] / (1/cos x . 1/sin x)
[(sin x . sin x + cos x . cos x) / (sin x . cos x)] (1/sin x cos x)
[(sin x . sin x + cos x . cos x) / (sin x . cos x)] (sin x . cos x)
then
sin x. sin x + cos x . cos x
sin2x+cos2x
=1
The answer is 1.
What else can I help you with?
What is the derivative of csc x?
The derivative of csc(x) is -cot(x)csc(x).
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
Cotangent squared plus one equals cosecant squared?
yes 1 + cot x^2 = csc x^2
How do you simplify sec x cot x?
sec(x)*cot(x) = (1/cos(x))*(cos(x)/sin(x)) = (1/sin(x)) = csc(x)
What is the derivative of cscx?
d/dx csc(x) = - csc(x) tan(x)
Csc divided by cot squared equals tan multiplied by sec?
Yes.
What is the derivative of csc x?
The derivative of csc(x) is -cot(x)csc(x).
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
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
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) ).
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)
What is the derivative of 3tanx-4cscx?
7
Tan plus cot divided by tan equals csc squared?
(tanx+cotx)/tanx=(tanx/tanx) + (cotx/tanx) = 1 + (cosx/sinx)/(sinx/cosx)=1 + cos2x/sin2x = 1+cot2x= csc2x This is a pythagorean identity.
What is the answer of csc x plus cot x?
Without an "equals" sign somewhere, no question has been asked,so there's nothing there that needs an answer.Is it the sum that you're looking for ?csc(x) + cot(x) = 1/sin(x) + cos(x)/sin(x) = [1 + cos(x)] / sin(x)
Express csc theta in terms of cot theta theta is in quadrant 3?
It is -sqrt(1 + cot^2 theta)
