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# How do you simplify csc theta cot theta cos theta?

###### Wiki User

cosec(q)*cot(q)*cos(q) = 1/sin(q)*cot(q)*cos(q) = cot2(q)

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## Related Questions

### How do you simplify csc theta -cot theta cos theta?

For a start, try converting everything to sines and cosines.

### How do you simplify cos theta times csc theta divided by tan theta?

'csc' = 1/sin'tan' = sin/cosSo it must follow that(cos) (csc) / (tan) = (cos) (1/sin)/(sin/cos) = (cos) (1/sin) (cos/sin) = (cos/sin)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.

### 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)

### How do you simplify sin theta times csc theta divided by tan theta?

Since sin(theta) = 1/cosec(theta) the first two terms simply camcel out and you are left with 1 divided by tan(theta), which is cot(theta).

### How do you simplify sec x cot x?

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

### How do you simplify csc theta tan theta?

With all due respect, you don't really want to know howto solve it.You just want the solution.csc(&Icirc;&tilde;) = 1/sin(&Icirc;&tilde;)tan(&Icirc;&tilde;) = sin(&Icirc;&tilde;)/cos(&Icirc;&tilde;)csc(&Icirc;&tilde;) x tan(&Icirc;&tilde;) = 1/sin(&Icirc;&tilde;) x sin(&Icirc;&tilde;)/cos(&Icirc;&tilde;) = 1/cos(&Icirc;&tilde;) = sec(&Icirc;&tilde;)

### How do i simplify csc theta divided by sec theta?

By converting cosecants and secants to the equivalent sine and cosine functions. For example, csc theta is the same as 1 / sin thetha.

### 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.

### 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 csc theta?

That depends on the value of the angle, theta. csc is short for "cosecans", and is the reciprocal of the sine. That is, csc theta = 1 / sin theta.

### 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.

### What is the exact value of cos theta if csc theta -4 with theta in quadrant III?

csc &theta; = 1/sin &theta; &rarr; sin &theta; = -1/4 cos&sup2; &theta; + sin&sup2; &theta; = 1 &rarr; cos &theta; = &plusmn; &radic;(1 - sin&sup2; &theta;) = &plusmn; &radic;(1 - &frac14;&sup2;) = &plusmn; &radic;(1- 1/16) = &plusmn; &radic;(15/16) = &plusmn; (&radic;15)/4 In Quadrant III both cos and sin are negative &rarr; cos &theta;= -(&radic;15)/4

### 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 figure this out Sin theta equals -0.0138 so theta equals what?

Depending on your calculator, you should have an arcsin function, which appears as sin^-1. It's usually a 2nd function of the sin key. If you don't have this function, there are many free calculators you can download... just google scientific calculator downloads.Anyway, this inverse function will give you theta when you plug in the value of sin theta. Here's the algebra written out:sin(theta)=-0.0138arcsin(sin(theta))=arcsin(-0.0138)theta=.......The inverse function applied to both sides of the equation "cancels out" the sin function and yields the value of the angle that was originally plugged into the function, in this case theta. You can use this principle to solve for theta for any of the other trig functions:arccos(cos(theta))=thetaarctan(tan(theta))=thetaand so on, but calculators usually only have these three inverse functions, so if you encounter a problem using sec, csc, or cot, you need to rewrite it as cos, sin, or tan.sec=1/coscsc=1/sincot=1/tan

### H ow do you verify that csc theta tan theta sec theta?

csc[]tan[] = sec[]. L: Change csc[] into one over sin[]. Change tan[] into sin[] over cos[]. R: Change sec[] into one over cos[]. 1/sin[] times sin[]/cos[] = 1/cos[]. L: To multiply 2 fractions, multiply the numerators, and multiply the denominators, and put the numerators' product over the denominators' product. R: Nothing more to do. sin[]/sin[]cos[] = 1/cos[]. L: You have a sin[] on both top and bottom. Cross them off to get a one on the top. 1/cos[] = 1/cos[]. Done. [] is theta. L is the left side of the equation. R is the right side.

### 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 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)

### 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)

### Cot 70 plus 4cos70 equals?

cot 70 + 4 cos 70 = cos 70 / sin 70 + 4 cos 70 = cos 70 (1/sin 70 + 4) = cos 70 (csc 70 + 4) Numerical answer varies, depending on whether 70 is in degrees, radians, or grads.

### What is the derivative of csc x?

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

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