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The answer depends on how the ratios are defined. In some cases tan is DEFINED as the ratio of sine and cosine rather than from the angle in a right angled triangle.
If the trig ratios were defined in terms of a right angled triangle, thensine is the ratio of the opposite side to the hypotenuse,
cosine is the ratio of the adjacent side to the hypotenuse,
and tangent is the ratio of the opposite side to the adjacent side.
It is then easy to see that sin/cos = (opp/hyp)/(adj/hyp) = opp/adj = tan.
If sine and cosine are defined as infinite sums for angles measured in radians, ie
sin(x) = x - x^3/3! + x^5/5! - x^7/7! + ...
and
cos = 1 - x^2/2! + x^4/4! - x^6/6! + ...
then it is less easy to see tan = sin/cos.
What else can I help you with?
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
What is the identity for tan theta?
The identity for tan(theta) is sin(theta)/cos(theta).
What is tan theta in terms of sin theta in quadrant II?
tan = sin/cos Now cos2 = 1 - sin2 so cos = +/- sqrt(1 - sin2) In the second quadrant, cos is negative, so cos = - sqrt(1 - sin2) So that tan = sin/[-sqrt(1 - sin2)] or -sin/sqrt(1 - sin2)
What is tan theta in terms of sin theta?
Almost by definition, tan θ = sin θ / cos θ You can convert this to sine θ in several ways, for example: sin θ / cos θ = sin θ / cos (pi/2 - θ) Or here is another way, using the Pythagorean identity: sin θ / cos θ = sin θ / root(1 - sin2θ)
How do you verify the identity of cos θ tan θ equals sin θ?
To show that (cos tan = sin) ??? Remember that tan = (sin/cos) When you substitute it for tan, cos tan = cos (sin/cos) = sin QED
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 tan theta cos theta?
Remember that tan = sin/cos. So your expression is sin/cos times cos. That's sin(theta).
What is the identity for tan theta?
The identity for tan(theta) is sin(theta)/cos(theta).
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 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).
What is tan theta in terms of sin theta in quadrant II?
tan = sin/cos Now cos2 = 1 - sin2 so cos = +/- sqrt(1 - sin2) In the second quadrant, cos is negative, so cos = - sqrt(1 - sin2) So that tan = sin/[-sqrt(1 - sin2)] or -sin/sqrt(1 - sin2)
What is the difference between sin cos and tan?
In a right triangle, the right angle is formed by sides a and b. Side c is the hypotenuse.Theta is the interior angle that joins (let's say) sides b and c. The sin of theta is the length of a over the length of c. The cos of theta is the length of b over the length of c. The tan of theta is the length of a over the length of b.Sin theta= opposite divided by hypotenuse. Cos theta=adjacent divided by hypotenuse. Tan theta=opposite over adjacent.(Sin1-Cos1)-Tan1=-1.25623905Sorry, that was a mathematician's joke.
What does cos divided by sin equal?
tan x
What is tan theta in terms of sin theta?
Almost by definition, tan θ = sin θ / cos θ You can convert this to sine θ in several ways, for example: sin θ / cos θ = sin θ / cos (pi/2 - θ) Or here is another way, using the Pythagorean identity: sin θ / cos θ = sin θ / root(1 - sin2θ)
What is tan squared theta minus sec squared theta simplified?
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
How do you verify 1 divided by cos to the second theta minus tan to the second theta equals cos to the second theta plus 1 divided by csc to the second theta?
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How do you work out the remaining side of a right angled triangle using theta and one side?
Suppose triangle ABC is right angled at C. Suppose you are given that the angle at B is theta. Thenif you know the length of AB (the hypotenuse), thenBC = AB*cos(theta) andAC = AB*sin(theta)if you know the length of BC, thenAB = BC/cos(theta) andAC = BC*tan(theta)if you know the length of AC, thenAB= AC/sin(theta) andBC = AC/tan(theta)
