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
0.75
The third quadrant.
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
Negative 1.047197551 etc, etc.
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
Cotan(theta) is the reciprocal of the tan(theta). So, cot(theta) = 1/2.
0.75
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)
-0.5736
If tan(theta) = x then sin(theta) = x/(sqrt(x2 + 1) so that csc(theta) = [(sqrt(x2 + 1)]/x = sqrt(1 + 1/x2)
The third quadrant.
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
tan theta = sqrt(2)/2 = 1/sqrt(2).
The identity for tan(theta) is sin(theta)/cos(theta).
Negative 1.047197551 etc, etc.