let's recall that the cosine/adjacent is negative on the II Quadrant, whilst the sine/opposite is positive on that same quadrant, also let's recall that the hypotenuse is never negative, since it's just a radius distance.
[tex]\bf cos(\theta )=\cfrac{\stackrel{adjacent}{-5}}{\stackrel{hypotenuse}{8}}\qquad \impliedby \textit{let's find the \underline{opposite side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-a^2}=b \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{8^2-(-5)^2}=b\implies \pm\sqrt{64-25}=b\implies \pm\sqrt{39}=b\implies \stackrel{II~Quadrant}{+\sqrt{39}=b} \\\\[-0.35em] ~\dotfill[/tex]
[tex]\bf csc(\theta )\implies \cfrac{\stackrel{hypotenuse}{8}}{\stackrel{opposite}{\sqrt{39}}}\implies \cfrac{8}{\sqrt{39}}\cdot \cfrac{\sqrt{39}}{\sqrt{39}}\implies \cfrac{8\sqrt{39}}{39} \\\\\\ cot(\theta )\implies \cfrac{\stackrel{adjacent}{-5}}{\stackrel{opposite}{\sqrt{39}}}\implies \cfrac{-5}{\sqrt{39}}\cdot \cfrac{\sqrt{39}}{\sqrt{39}}\implies \cfrac{-5\sqrt{39}}{39}[/tex]
The exact value of cosec(O) and cot(O) are (8 / √39) and (-5 / √39).
In 1st quadrant - (All trigonometric ratio are +ve)
In 2nd quadrant - ( Only Sin(O) , Cosec(O) are +ve)
In 3rd quadrant - ( Only Tan(O) , Cot(O) are +ve)
In 4th quadrant - ( Only Cos(O) , Sec(O) are +ve)
The square of the hypotenuse of a right angled triangle is equal to the sum of the squares of the other two sides of the right angled triangle.
Here P , B , H are referred to as Perpendicular, Base and Hypotenuse of a triangle respectively.
Cos(O) = B / H
Cos (O) = - 5 / 8
Hence B = 5 , H = 8
Applying Pythagoras theorem:
H^2 = P^2 + B^2
(8)^2 = P^2 + (5)^2
64 = P^2 + 25
P^2 = 39
P = √39
As the calculation has to be performed in the 2nd quadrant, the value of cosec(O) is +ve and the value of cot(O) is -ve.
Cosec(O) = H / P = 8 / √39
Cot(O) = B / P = (-5 )/ √39
Hence the value of cosec(O) and cot(O) are 8 / √39 and (-5 )/ √39 respectively.
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