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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 are the graphs of sec x and csc x related?
They are co-functions meaning that 90 - sec x = csc x.
Given tan Theta equals negative 15 divided by 8 and 90 degrees is less than or equal to theta which is less than or equals to 180 state 5 other trigonometric ratios and determine the measure of theta?
sin(theta) = 15/17, cosec(theta) = 17/15 cos(theta) = -8/17, sec(theta) = -17/8 cotan(theta) = -8/15 theta = 2.0608 radians.
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.
What is sec squared theta times cos squared theta minus tan squared theta?
Sec^(2) = 1 / Cos^(2) Hence Sec^(2) X Cos^(2) = 1/Cos^(2) X Cos^(2) Cancel down by ' Cos^(2) '. 1/1 X 1 = 1 1 - Tan^(2) (The ANSWER) !!!!! Factor ( 1 - Tan)(1 + Tan)
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.
Csc divided by cot squared equals tan multiplied by sec?
Yes.
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.
What is whole under root of sec theta - 1 over sec theta 1?
Ut is equual to tan(theta) / (sec(theta) + 1)
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 sec x cot x?
sec(x)*cot(x) = (1/cos(x))*(cos(x)/sin(x)) = (1/sin(x)) = csc(x)
How are the graphs of sec x and csc x related?
They are co-functions meaning that 90 - sec x = csc x.
What is sec of 2 theta?
The secant of an angle (2\theta), denoted as (\sec(2\theta)), is the reciprocal of the cosine of that angle. It can be expressed mathematically as (\sec(2\theta) = \frac{1}{\cos(2\theta)}). The value of (\sec(2\theta)) will depend on the specific angle (2\theta) and can be found using trigonometric identities or a calculator.
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 tan x csc x?
tan(x)*csc(x) = sec(x)
How do you simplify sec²x minus 1 divided by sec²x?
Use your identity to write the numerator in terms of tangent.
