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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 cos theta?
cosec(q)*cot(q)*cos(q) = 1/sin(q)*cot(q)*cos(q) = cot2(q)
How do you simplify csc theta -cot theta cos theta?
For a start, try converting everything to sines and cosines.
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 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.
What is the exact value of cos theta if csc theta -4 with theta in quadrant III?
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
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.
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(Θ) = 1/sin(Θ)tan(Θ) = sin(Θ)/cos(Θ)csc(Θ) x tan(Θ) = 1/sin(Θ) x sin(Θ)/cos(Θ) = 1/cos(Θ) = sec(Θ)
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 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 solve theta if cos squared theta equals 1 and 0 is less than or equal to theta which is less than 2pi?
cos2(theta) = 1 so cos(theta) = ±1 cos(theta) = -1 => theta = pi cos(theta) = 1 => theta = 0
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 tan theta cos theta?
Remember that tan = sin/cos. So your expression is sin/cos times cos. That's sin(theta).
How do you get the csc theta given tan theta in quadrant 1?
If tan(theta) = x then sin(theta) = x/(sqrt(x2 + 1) so that csc(theta) = [(sqrt(x2 + 1)]/x = sqrt(1 + 1/x2)
How do you verify 1 divided by cos to the second theta minus tan to the second theta equals cos to the second theta plus 1 divided by csc to the second theta?
Verify the identity:1/(cos θ)^2 - (tan θ)^2 = (cos θ)^2 + 1/(csc θ)^21/(cos θ)^2 - (sin θ)^2/(cos θ)^2 = (cos θ)^2 + (sin θ)^2 ?1 - (sin θ)^2/(cos θ)^2 = (cos θ)^2 + (sin θ)^2 ?(cos θ)^2/(cos θ)^2 = 1 ?1 = 1 TrueMethod 21/(cos θ)2 - (tan θ)2 =? (cos θ)2 + 1/(cscθ)21/(cos θ)2-(sinθ)2/(cos θ)2=? (cosθ)2+ sin(θ)21/(cos θ)2[1-sin(θ)2]=? cos(θ)2+sin(θ)21/cos(θ)2(cos(θ)2)=? 11=1 True
Express csc theta in terms of cot theta theta is in quadrant 3?
It is -sqrt(1 + cot^2 theta)
How do you find the area of a rectangle from diagonal and 1 angle?
The diagonal multiplied by sin(angle) gives one side of the rectangle and the diagonal times cos(theta) gives the other. So the area is (diagonal)2 x cos(theta) x sin(theta).
De Morgan's law in complex number?
(Sin theta + cos theta)^n= sin n theta + cos n theta
What is sec theta - 1 over sec theta?
Let 'theta' = A [as 'A' is easier to type] sec A - 1/(sec A) = 1/(cos A) - cos A = (1 - cos^2 A)/(cos A) = (sin^2 A)/(cos A) = (tan A)*(sin A) Then you can swap back the 'A' with theta
What is Cos Theta minus Cos Theta?
Zero. Anything minus itself is zero.
How do you solve cos theta subtract cos squared theta divide 1 plus cos theta?
The question contains an expression but not an equation. An expression cannot be solved.
If cos and theta 0.65 what is the value of sin and theta?
You can use the Pythagorean identity to solve this:(sin theta) squared + (cos theta) squared = 1.
In which quardant are the terminal arms of the angle lie when sin thita is less then zero and cos thita is greater then zero?
The fourth Across the quadrants sin theta and cos theta vary: sin theta: + + - - cos theta: + - - + So for sin theta < 0, it's the third or fourth quadrant And for cos theta > 0 , it's the first or fourth quadrant. So for sin theta < 0 and cos theta > 0 it's the fourth quadrant
What is cos 90 minus theta?
cosine (90- theta) = sine (theta)
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
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