No.
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
cos2(theta) = 1 cos2(theta) + sin2(theta) = 1 so sin2(theta) = 0 cos(2*theta) = cos2(theta) - sin2(theta) = 1 - 0 = 1
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
If tan(theta) = x then sin(theta) = x/(sqrt(x2 + 1) so that csc(theta) = [(sqrt(x2 + 1)]/x = sqrt(1 + 1/x2)
Ut is equual to tan(theta) / (sec(theta) + 1)
zero
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.
No.
By converting cosecants and secants to the equivalent sine and cosine functions. For example, csc theta is the same as 1 / sin thetha.
It also equals 13 12.
Tan^2
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
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