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Q: How do you get the csc theta given tan theta in quadrant 1?

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Cotan(theta) is the reciprocal of the tan(theta). So, cot(theta) = 1/2.

'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

0.75

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).

tan2(theta) + 5*tan(theta) = 0 => tan(theta)*[tan(theta) + 5] = 0=> tan(theta) = 0 or tan(theta) = -5If tan(theta) = 0 then tan(theta) + cot(theta) is not defined.If tan(theta) = -5 then tan(theta) + cot(theta) = -5 - 1/5 = -5.2

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.

No, they cannot all be negative and retain the same value for theta, as is shown with the four quadrants and their trigonemtric properties. For example, in the first quadrant (0

tan(x)*csc(x) = sec(x)

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)

If tan A = 1/2, then sin A = ? We use the Pythagorean identity 1 + cot2 A = csc2 A to find csc A, and then the reciprocal identity sin A = 1/csc A to find sin A. tan A = 1/2 (since tan A is positive, A is in the first or the third quadrant) cot A = 1/tan A = 1/(1/2) = 2 1 + cot2 A = csc2 A 1 + (2)2 = csc2 A 5 = csc2 A √5 = csc A (when A is in the first quadrant) 1/√5 = sin A √5/5 = sin A If A is in the third quadrant, then sin A = -√5/5.

it is POSITIVE because tangent is said to be as OPPOSITE all over ADJACENT side of the triangle. since the opposite and adjacent sides of theta in Quadrant 3 are both negative, the quotient of two negative integers is POSITIVE. in third quadrant tanƟ= -O/-A

tan theta = sqrt(2)/2 = 1/sqrt(2).

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