Find the exact value of csc theta for an angle theta with cosine theta with equals -2/5 and with its terminal side in quadrant 2

A

Step-by-step explanation:

Cos(theta) = -2/5

From sin^2(theta) + cos^2(theta) = 1 we get

sin^2(theta) = 1 - cos^2(theta)

sin(theta) = sqrt(1 - cos^2(theta)). The root is the positive one in quad 2

sin(theta) = sqrt(1 - cos^2(theta))

sin(theta) = sqrt(1 - (-2/5)^2)

sin(theta) = sqrt(1 - 4/25)

sin(theta) = sqrt(21/25)

sin(theta) = sqrt(21)/5

csc(theta) = 1/sin(theta)

csc(theta) = 1/sqrt(21)/5

csc(theta) = 5/sqrt(21)

The answer should be the denominator rationalized as in

csc(theta) = 5 * sqrt(21) / 21.

However that is not an option, so the answer is A

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step-by-step explanation:

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central angle of a triangle has the same measure as its intercepted arc.

step-by-step explanation: