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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.
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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
