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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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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 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 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.
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 cot of theta times sin of theta?
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.
How do you simplify csc theta -cot theta cos theta?
For a start, try converting everything to sines and cosines.
How do you simplify csc theta cot theta cos theta?
cosec(q)*cot(q)*cos(q) = 1/sin(q)*cot(q)*cos(q) = cot2(q)
Express csc theta in terms of cot theta theta is in quadrant 3?
It is -sqrt(1 + cot^2 theta)
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).
Can you simplify 1-cot x?
csc^2x+cot^2x=1
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.
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
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 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)
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)
What is cos theta multiplied by csc theta?
It is cotangent(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)
