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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.

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0It is cotangent(theta).

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

If tan(theta) = x then sin(theta) = x/(sqrt(x2 + 1) so that csc(theta) = [(sqrt(x2 + 1)]/x = sqrt(1 + 1/x2)

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)

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

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

'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

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).

sin theta and csc theta are reciprocal functions because sin = y/r and csc = r/y you use the same 2 sides of a triangle, but you use the reciprocal.

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

csc θ = 1/sin θ → sin θ = -1/4 cos² θ + sin² θ = 1 → cos θ = ± √(1 - sin² θ) = ± √(1 - ¼²) = ± √(1- 1/16) = ± √(15/16) = ± (√15)/4 In Quadrant III both cos and sin are negative → cos θ= -(√15)/4

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

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

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.

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)

Sine Theta (sin θ) = opposite/hypotenuse = a/c Cosine Theta (cos θ) = adjacent/hypotenuse = b/c Tangent Theta (tan θ) = opposite/adjacent = a/b Cotangent Theta (cot θ) = adjacent/opposite = b/a Secant Theta (sec θ) = hypotenuse/adjacent = c/b Cosecant Theta (csc θ) = hypotenuse/opposite = c/a You may need to look on the link below for some sample calculations

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.

csc(7 degrees) = 8.2055 csc(7 radians) = 1.5221 csc(7 grads) = 9.1129

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

Below link may help.You need to memorize the basic formulas and it will be easy.pythagorean theoremc² = a² + b²Sine Theta (sin θ) = opposite/hypotenuse = a/cCosine Theta (cos θ) = adjacent/hypotenuse = b/cTangent Theta (tan θ) = opposite/adjacent = a/bCotangent Theta (cot θ) = adjacent/opposite = b/aSecant Theta (sec θ) = hypotenuse/adjacent = c/bCosecant Theta (csc θ) = hypotenuse/opposite = c/a

When you subtract theta from 180 ( if theta is between 90 degrees and 180 degrees) you will get the reference angle of theta; the results of sine theta and sine of its reference angle will be the same and only the sign will be different depends on which quadrant the angle is located. Ex. 150 degrees' reference angle will be 30 degrees (180-150) sin150=1/2 (2nd quadrant); sin30=1/2 (1st quadrant) 1st quadrant: all trig functions are positive 2nd: sine and csc are positive 3rd: tangent and cot are positive 4th: cosine and secant are positive

look yo ugly one csc is bien csc lol call....786 487 6958

CSC India was created in 1991.

The population of CSC India is 20,000.

cos2(theta) = 1 cos2(theta) + sin2(theta) = 1 so sin2(theta) = 0 cos(2*theta) = cos2(theta) - sin2(theta) = 1 - 0 = 1

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