If cot (∅) = 5/2 and cos (∅) < 0, then what are the exact values of
tan (∅) and csc (∅)
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cot (∅) = 5/2
tan(∅)=1/cot(∅)....= 2/5
or cot(∅) =base/perpendicular line of a triangle(right angled triangle)
applying pythoghoras theorem
√(-5)² +(-2)² = √29
csc(∅)=hypotenuse/perpendicular
=√29 /2
tan(∅)=1/cot(∅)....= 2/5
or cot(∅) =base/perpendicular line of a triangle(right angled triangle)
applying pythoghoras theorem
√(-5)² +(-2)² = √29
csc(∅)=hypotenuse/perpendicular
=√29 /2
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Solutions
cot(∅) = x/y = 5/2 or -5/-2
Because the hypotenuse is always a positive value, the value of "y" must be a negative value in order for the cosine to be less than zero. We must then conclude that in our right triangle "x" and "y" would both be negatives, since the tangent is positive.
√(-5)² (-2)² = √29 = hyp
csc(∅)=hypotenuse/perpendicular
csc(∅) = hyp/y = 29/-2 = - 29/-2
cot(∅) = x/y = 5/2 or -5/-2
Because the hypotenuse is always a positive value, the value of "y" must be a negative value in order for the cosine to be less than zero. We must then conclude that in our right triangle "x" and "y" would both be negatives, since the tangent is positive.
√(-5)² (-2)² = √29 = hyp
csc(∅)=hypotenuse/perpendicular
csc(∅) = hyp/y = 29/-2 = - 29/-2
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