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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.

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Q: H ow do you verify that csc theta tan theta sec theta?
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What is tan x csc x?

tan(x)*csc(x) = sec(x)


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)


What is whole under root of sec theta - 1 over sec theta 1?

Ut is equual to tan(theta) / (sec(theta) + 1)


What is sec squared theta times cos squared theta minus tan squared theta?

Tan^2


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 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).


Sec squared theta plus tan squared theta equals to 13 12 and sec raised to 4 theta minus tan raised to 4 theta equals to how much?

It also equals 13 12.


How can you verify 1 plus tan theta divided by 1 minus tan theta equals cot theta plus 1 divided by cot theta minus 1?

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


Csc divided by cot squared equals tan multiplied by sec?

Yes.


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


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)


Csc x tan x?

If you want to simplify that, it usually helps to express all the trigonometric functions in terms of sines and cosines.


If tansqtheta plus 5tantheta0 find the value of tantheta plus cottheta?

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


What is sec theta if tan theta equals 2 with theta in quadrant 3?

If tan theta equals 2, then the sides of the triangle could be -2, -1, and square root of 5 (I used the Pythagorean Theorem to get this). From this, sec theta is negative square root of 5. It is negative because theta is in the third quadrant, where cosine, secant, sine, and cosecant are all negative.


How do you convert slope percent into tan degree value?

Let m be the slope in percent and theta be the angle in question. tan (theta) = m/100 theta = arctan (m/100) To verify the result, we know the following: tan 0 = 0 tan (45 degrees) = 1 = 100% tan (90 degrees) = infinity For example, if 0 < m < 100%, then 0 < theta < 45 degrees.


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.


When is tan theta theta?

tan (theta x theta) : must square the value of the angle, theta, before applying the trig function, tangent.


What isTan theta?

Tan theta is a function of the number theta.


What is theta when tantheta plus sectheta equals 1 please show work?

copy this and paste in your browsers address window http://www.wolframalpha.com/input/?i=tan+theta+%2B+sec+theta+%3D1


What is the second derivative of ln(tan(x))?

f'(x) = 1/tan(x) * sec^2(x) where * means multiply and ^ means to the power of. = cot(x) * sec^2(x) f''(x) = f'(cot(x)*sec^2(x) + cot(x)*f'[sec^2(x)] = -csc^2(x)*sec^2(x) + cot(x)*2tan(x)sec^2(x) = sec^2(x) [cot(x)-csc^2(x)] +2tan(x)cot(x) = sec^2(x) [cot(x)-csc^2(x)] +2


If tan Theta equals 2 with Theta in Quadrant 3 find cot Theta?

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


Any interesting way in teaching right angle trigonometry?

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


Values of the 6 trigonometric functions?

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


What is tangent of theta as a percentage?

It is 100*tan(theta).


What is sec squared theta minus tan squared theta?

1 - sin2(q) = cos2(q)dividing through by cos2(q),sec2(q) - tan2(q) = 1