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Answered 2011-02-21 14:10:09
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
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Ut is equual to tan(theta) / (sec(theta) + 1)
For such simplifications, it is usually convenient to convert any trigonometric function that is not sine or cosine, into sine or cosine. In this case, you have: sin theta / sec theta = sin theta / (1/cos theta) = sin theta cos 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.
By converting cosecants and secants to the equivalent sine and cosine functions. For example, csc theta is the same as 1 / sin thetha.
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.
copy this and paste in your browsers address window http://www.wolframalpha.com/input/?i=tan+theta+%2B+sec+theta+%3D1
1 - sin2(q) = cos2(q)dividing through by cos2(q),sec2(q) - tan2(q) = 1
There is none because it is not true.
cos2(theta) = 1 cos2(theta) + sin2(theta) = 1 so sin2(theta) = 0 cos(2*theta) = cos2(theta) - sin2(theta) = 1 - 0 = 1
cos2(theta) = 1 so cos(theta) = ±1 cos(theta) = -1 => theta = pi cos(theta) = 1 => theta = 0
There are three of them. Granted this means that there are different variations of all three. I'll show you the variations as well. This is coming straight from my Math 1060 (Trigonometry) notebook. Sorry there is no key to represent the angle; Theta.1. Sin2 (of Theta) + Cos2 (of Theta)= 1Variations: Sin2 (of Theta) = 1- Cos2 (of Theta)AND: Cos2 (of Theta) = 1-Sin2 (of Theta)2. Tan2 (of Theta) + 1 = sec2 (of Theta)Variations: Tan2 (of Theta) = Sec2 (of Theta) -13. 1 + Cot2 (of Theta) = Csc2 (of Theta)Variations: Cot2 (of Theta) = Csc2 (of Theta) -1
4*cos2(theta) = 1 cos2(theta) = 1/4 cos(theta) = sqrt(1/4) = ±1/2 Now cos(theta) = 1/2 => theta = 60 + 360k or theta = 300 + 360k while Now cos(theta) = -1/2 => theta = 120 + 360k or theta = 240 + 360k where k is an integer.
sin(t) = 2/3 sin2(t) + cos2(t) = 1 so cos(t) = ± sqrt[1 - sin2(t)] but because t is in the first quadrant, cos(t) > 0 so cos(t) = + sqrt[1 - sin2(t)] = sqrt[1 - 4/9] = sqrt[5/9] = sqrt(5)/3 Then sec(t) = 1/cos(t) = 1/sqrt(5)/3 = 3/sqrt(5) = 3*sqrt(5)/5
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
-1 < sine(theta) < 1 so sine(theta) cannot be 3125
It depends if 1 plus tan theta is divided or multiplied by 1 minus tan theta.
Cosine squared theta = 1 + Sine squared theta
If tan(theta) = x then sin(theta) = x/(sqrt(x2 + 1) so that csc(theta) = [(sqrt(x2 + 1)]/x = sqrt(1 + 1/x2)
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).
sin(theta) = 15/17, cosec(theta) = 17/15 cos(theta) = -8/17, sec(theta) = -17/8 cotan(theta) = -8/15 theta = 2.0608 radians.
To determine what negative sine squared plus cosine squared is equal to, start with the primary trigonometric identity, which is based on the pythagorean theorem...sin2(theta) + cos2(theta) = 1... and then solve for the question...cos2(theta) = 1 - sin2(theta)2 cos2(theta) = 1 - sin2(theta) + cos2(theta)2 cos2(theta) - 1 = - sin2(theta) + cos2(theta)
2 sin^2 theta = 1/4 sin^2 theta = 1/8 sin theta = sqrt(1/8) theta = arcsin(sqrt(1/8))
It's 1/2 of sin(2 theta) .
Cotan(theta) is the reciprocal of the tan(theta). So, cot(theta) = 1/2.
It is -sqrt(1 + cot^2 theta)
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