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Given: sin(a) = 5/13, π/2 < A < π and tan(b) = -√(π/2), π/2 < B < π. What is tan(A-B)?

A) -5/√(13)
B) 5√(13)
C) -√(13)
D) √(13)

User Kkgarg
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1 Answer

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Final answer:

To find tan(A-B), we first find the values of cos(a) and cos(b) using the given sin(a) and tan(b). Then, we use the identities tan(A-B) = (tan(A) - tan(B))/(1 + tan(A)tan(B)) to evaluate tan(A-B). The final answer is -5/sqrt(13).

Step-by-step explanation:

To find the value of tan(A-B), we need to use the trigonometric identities and the given values of sin(a) and tan(b). First, let's find the values of cos(a) and cos(b). Since sin(a) = 5/13, we can use the Pythagorean identity sin^2(a) + cos^2(a) = 1 to find cos(a). So, cos^2(a) = 1 - sin^2(a) = 1 - (5/13)^2 = 1 - 25/169 = 144/169. Taking the square root of both sides gives cos(a) = sqrt(144/169) = 12/13.

Next, since tan(b) = -sqrt(pi/2), we can use the relationship tan(b) = sin(b)/cos(b) to find sin(b) and cos(b). We know that sin^2(b) + cos^2(b) = 1, so sin^2(b) + (sin(b)/tan(b))^2 = 1. Substituting the given value of tan(b), we get sin^2(b) + (sin(b)/(-sqrt(pi/2)))^2 = 1. Simplifying this equation, we find sin^2(b) + (sin^2(b) * 2/pi) = 1. Solving for sin(b), we find sin(b) = sqrt(pi/(2+pi)). Since sin(b) > 0 and cos(b) = sqrt(1 - sin^2(b)), we have cos(b) = sqrt(1 - (pi/(2+pi)) = sqrt((2+pi-pi)/(2+pi)) = sqrt(2/(2+pi)).

Now, we can evaluate tan(A-B) using the identities tan(A-B) = (tan(A) - tan(B))/(1 + tan(A)tan(B)). Substituting the values of sin(a), cos(a), sin(b), and cos(b) into the formula, we have tan(A-B) = (sin(a)/cos(a) - sin(b)/cos(b))/(1 + (sin(a)/cos(a))(sin(b)/cos(b))). Simplifying this expression, we get tan(A-B) = (((5/13) / (12/13)) - (sqrt(pi/(2+pi)) / sqrt(2/(2+pi)))) / (1 + ((5/13) / (12/13)) * (sqrt(pi/(2+pi)) / sqrt(2/(2+pi)))). This can be further simplified to tan(A-B) = (5 - sqrt(pi(2+pi))/(12 + 5(sqrt(pi(2+pi)))) = -5/sqrt(13).

User Mark Notton
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