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No, I'm saying that cotangent of an angle in the 2nd quadrant must be negative.
well my answer was negative, but the teacher circled the positive one...
If theta is is quadrant II, the cosine is negative and sine is positive. Thus, tanget is negative and so cotangent must be negative.
So your saying that you take the answer I got which is -5/12 and add another negative since in quadrant 2, tan is negative, and you get 5/12?
On a math test, the question was as follows: If Cosθ = -5/13 and θ is in quadrant 2, determine Cotθ. I got -5/12, but the teacher circled 5/12 as the right answer.
Yes, it is the quadratic formula. If we let x = cos θ, then
which may be solved by means of the quadratic formula.
Oh yeah, how stupid of me
can you please explain how you got from this to this:
2cos²+cos-2 to cos = -1±√1-(4)(2)(-2)/2(2)
√ 2 sin²-cos=0
T # 8
If x*sin60°*tan30°=(sin45°*cot45°)/(cos30°*cosec60°), find x.
for this i got every part easily cept sin 45 because it equals a long decimal so it didnt seem right but i got 1.414
T # 7
Solve:-
Cos²θ/(Cot²θ-Cos²θ) = 3.
Not too sure how to do this but i manipulated the question until i got down to 4cos sin = 3 and then sin ≠ 3 and cos =3/4 I just dont know how to find this without a calculator.
does anyone know how to get this question? I got pi/2 and 3pi/3.
√2 sin² -sin = 0 Solve for values of sin. 0<x<2pi
does anyone know how to get this question? I got pi/2 and 3pi/3.
√2 sin² -sin = 0 Solve for values of sin. 0<x<2pi