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csc(x) + cot(x) = 1/sin(x) + cos(x)/sin(x) = [1 + cos(x)] / sin(x)

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โˆ™ 2010-10-26 05:03:10
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A polynomial of degree zero is a constant term

The grouping method of factoring can still be used when only some of the terms share a common factor A True B False

The sum or difference of p and q is the of the x-term in the trinomial

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Q: What is the answer of csc x plus cot x?
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What is the derivative of csc x?

The derivative of csc(x) is -cot(x)csc(x).

Can you simplify 1-cot x?


What is the derivative of 3tanx-4cscx?


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.

Cotangent squared plus one equals cosecant squared?

yes 1 + cot x^2 = csc x^2

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

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 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 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.

How do you simplify sec x cot x?

sec(x)*cot(x) = (1/cos(x))*(cos(x)/sin(x)) = (1/sin(x)) = csc(x)

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)

Verify that Cos theta cot theta plus sin theta equals csc theta?

It's easiest to show all of the work (explanations/identities), and x represents theta. cosxcotx + sinx = cscx cosx times cosx/sinx + sinx = csc x (Quotient Identity) cosx2 /sinx + sinx = csc x (multiplied) 1-sinx2/sinx + sinx = csc x (Pythagorean Identity) 1/sinx - sinx2/sinx + sinx = csc x (seperate fraction) 1/sinx -sinx + sinx = csc x (canceled) 1/sinx = csc x (cancelled) csc x =csc x (Reciprocal Identity)

Are shifts sin cos tan equal to csc sec cot respectively?

No, they are the inverse functions, while csc, sec and cot are the reciprocal functions. To illustrate the difference, the inverse of f(x) = x+3 is f-1(x) = x-3 But the reciprocal of f(x) is 1/f(x) = 1/(x+3)

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.

Solution for tan x plus cot x divided by sec x csc x?

(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 is sin cos tan csc sec cot of 135 degrees?

All those can be calculated quickly with your calculator. Just be sure it is in "degrees" mode (not in radians). Also, use the following identities: csc(x) = 1 / sin(x) sec(x) = 1 / cos(x) cot(x) = 1 / tan(x) or the equivalent cos(x) / sin(x)

How do you integrate the cscnx?

The integral for csc(u)dx is -ln|csc(u) + cot(u)| + C.

How do you simplify csc theta minus cot x theta times cos theta plus 1?

There can be no significant simplicfication if some of the angles are theta and others are x, so assume that all angles are x. [csc(x) - cot(x)]*[cos(x) + 1] =[1/sin(x) - cos(x)/sin(x)]*[cos(x) + 1] =1/sin(x)*[1 - cos(x)]*[cos(x) + 1] =1/sin(x)*[1 - cos2(x)] =1/sin(x)*[sin2(x)] = sin(x)

What is cot x sin x simplified?

To simplify such expressions, it helps to express all trigonometric functions in terms of sines and cosines. That is, convert tan, cot, sec or csc to their equivalent in terms of sin and cos.

What is tan x csc x?

tan(x)*csc(x) = sec(x)

Express csc theta in terms of cot theta theta is in quadrant 3?

It is -sqrt(1 + cot^2 theta)

Csc divided by cot squared equals tan multiplied by sec?


What is the derivative of cscx?

d/dx csc(x) = - csc(x) tan(x)

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