answersLogoWhite

0

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)

User Avatar

Wiki User

15y ago

What else can I help you with?

Related Questions

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


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 simplify csc theta cot theta?

There are 6 basic trig functions.sin(x) = 1/csc(x)cos(x) = 1/sec(x)tan(x) = sin(x)/cos(x) or 1/cot(x)csc(x) = 1/sin(x)sec(x) = 1/cos(x)cot(x) = cos(x)/sin(x) or 1/tan(x)---- In your problem csc(x)*cot(x) we can simplify csc(x).csc(x) = 1/sin(x)Similarly, cot(x) = cos(x)/sin(x).csc(x)*cot(x) = (1/sin[x])*(cos[x]/sin[x])= cos(x)/sin2(x) = cos(x) * 1/sin2(x)Either of the above answers should work.In general, try converting your trig functions into sine and cosine to make things simpler.


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


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)


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 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)


How do you make 'cot' and 'csc' on a TI-84 graphing calculator?

From math class, some trigonometric identities: cot x = 1/tan x csc x = 1/sin x sec x = 1/cos x There are no built-in cot or csc formulas, so use the above. Remember that these give errors when tan x, sin x, or cos x are equal to 0.


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)