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How do you simplify csc theta tan theta?
With all due respect, you don't really want to know howto solve it.You just want the solution.csc(Θ) = 1/sin(Θ)tan(Θ) = sin(Θ)/cos(Θ)csc(Θ) x tan(Θ) = 1/sin(Θ) x sin(Θ)/cos(Θ) = 1/cos(Θ) = sec(Θ)
Given sin theta equals 7 over 13 and tan theta is less than 0 find tan theta?
sin(t) = 7/13 cos2(t) = 1 - sin2(t) = (169 - 49)/169 = 120/169 so cos(t) = ±sqrt(120)/13. But sin(t) > 0, tan(t) < 0 implies t is in the second quadrant so cos(t) = -sqrt(120)/13 And then tan(t) = sin(t)/cos(t) = -7/sqrt(120) = -0.6390 (approx).
What is tan theta minus cot theta?
-2(cot2theta)
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 are both solutions between -180 and 180 degrees of the equation (tan theta -7over10)?
They are theta = -34.99 degrees and 145.09 deg.
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
If tan Theta equals 2 with Theta in Quadrant 3 find cot Theta?
Cotan(theta) is the reciprocal of the tan(theta). So, cot(theta) = 1/2.
If sin theta equals 3/4 and theta is in quadrant II what is the value of tan theta?
0.75
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 sec theta if tan theta equals 2 with theta in quadrant 3?
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.
Is tan theta in the 3rd quadrant negative?
In the third quadrant, both the x and y coordinates are negative. Since tangent is defined as the ratio of the opposite side to the adjacent side in a right triangle, in the third quadrant where both sides are negative, the tangent of an angle theta will be positive. Therefore, tan theta is not negative in the third quadrant.
By using trigonometric identities find the value of sin A if tan A equals a half?
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.
If tansqtheta plus 5tantheta0 find the value of tantheta plus cottheta?
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
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)
Is it possible for sin theta cos theta and tan theta to all be negative for the same value of theta?
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
How do you simplify csc theta tan theta?
With all due respect, you don't really want to know howto solve it.You just want the solution.csc(Θ) = 1/sin(Θ)tan(Θ) = sin(Θ)/cos(Θ)csc(Θ) x tan(Θ) = 1/sin(Θ) x sin(Θ)/cos(Θ) = 1/cos(Θ) = sec(Θ)
How do you find tan-theta when sin-theta equals -0.5736 and cos-theta is greater than 0?
-0.5736
