zero
sec(x)=1/cos(x), by definition of secant.
cos(t) - cos(t)*sin2(t) = cos(t)*[1 - sin2(t)] But [1 - sin2(t)] = cos2(t) So, the expression = cos(t)*cos2(t) = cos3(t)
cosec(q)*cot(q)*cos(q) = 1/sin(q)*cot(q)*cos(q) = cot2(q)
For any calculator Sec(Secant) = 1/Cos Csc (Cosecant) = 1/ Sin Cot (Cotangent) = 1/Tan
Remember ' Csc(Cosecant)' = 1 /sin Hence Cos X 1/Sin = Cos/Sin = Cot (Cotangent) Do NOT confuse with Tan = Sin /Cos. Notice the positions of trig. functions in the ratios.
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
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)
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.
Ut is equual to tan(theta) / (sec(theta) + 1)
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.
tan θ = sin θ / cos θ sec θ = 1 / cos θ sin ² θ + cos² θ = 1 → sin² θ - 1 = - cos² θ → tan² θ - sec² θ = (sin θ / cos θ)² - (1 / cos θ)² = sin² θ / cos² θ - 1 / cos² θ = (sin² θ - 1) / cos² θ = - cos² θ / cos² θ = -1
cos2(theta) = 1 so cos(theta) = ±1 cos(theta) = -1 => theta = pi cos(theta) = 1 => theta = 0
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
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)
1/cos y = sec y
'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