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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.

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Q: How do you simplify csc theta cot theta?

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

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

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

csc^2x+cot^2x=1

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

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.

'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

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 cotangent(theta).

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)

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

Cotan(theta) is the reciprocal of the tan(theta). So, cot(theta) = 1/2.

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

Yes, it is.

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

-2(cot2theta)

∫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

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.

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

It depends if 1 plus tan theta is divided or multiplied by 1 minus tan theta.

cot theta=tan(90-tetha)

tan2(theta) + 5*tan(theta) = 0 => tan(theta)*[tan(theta) + 5] = 0=> tan(theta) = 0 or tan(theta) = -5If tan(theta) = 0 then tan(theta) + cot(theta) is not defined.If tan(theta) = -5 then tan(theta) + cot(theta) = -5 - 1/5 = -5.2

By converting everything to sines and cosines. Since tan x = sin x / cos x, in the cotangent, which is the reciprocal of the tangent: cot x = cos x / sin x. You can replace any other variable (like thetha) for the angle.

whats the big doubt,cot/tan+1= 1+1= 2