find the exact value of the expression using the provided information. Find Tan (S+T) given that cos s=1/3 with s in quad I and sin T= -1/2 with T in quad IV

s is in Q1 (sin>0, cos>0)

so
cos(s)=1/3,
sin(s)=+√(1-(1/3)²)
=√(8/9)
tan(s)=√(8/9) / (1/3)
=√8

t is in Q4 (sin<0, cos>0)
sin(t)=-1/2
cos(t)=+√(1-(1/2)²)
=√(3/4)
tan(t)=(-1/2)/√(3/4)
=-1/√3
[rationalize by multiplying numerator and denominator by √3]
=-(√3)/3

Use the identity
tan(s+t)=(tan(s)+tan(t))/(1-tan(s)tan(t)) to calculate tan(s+t)
I get about 0.85.

To find the exact value of tan(S+T), we can use two trigonometric identities:

1) tan(A + B) = (tanA + tanB) / (1 - tanA*tanB)
2) tan(-A) = -tan(A)

Let's first find the values of tan S and tan T using the given information:

Given that cos S = 1/3 and S is in quad I, we can use the Pythagorean identity: cos^2 S + sin^2 S = 1, to find sin S.

sin S = sqrt(1 - cos^2 S)
= sqrt(1 - (1/3)^2)
= sqrt(1 - 1/9)
= sqrt(8/9)
= sqrt(8) / 3

So, tan S = sin S / cos S = (sqrt(8) / 3) / (1/3) = sqrt(8).

Given that sin T = -1/2 and T is in quad IV, we know that sin T is negative, and cos T = sqrt(1 - sin^2 T) = sqrt(1 - (1/2)^2) = sqrt(3/4) = sqrt(3) / 2.

So, tan T = sin T / cos T = (-1/2) / (sqrt(3) / 2) = -1 / sqrt(3) = -sqrt(3) / 3.

Now let's use the tan(A + B) identity to find tan(S + T):

tan(S + T) = (tan S + tan T) / (1 - tan S * tan T)
= (sqrt(8) + (-sqrt(3)/3)) / (1 - (sqrt(8) * (-sqrt(3)/3)))

To simplify, let's rationalize the denominator:

tan(S + T) = (sqrt(8) + (-sqrt(3)/3)) / (1 + sqrt(8)*sqrt(3) / 3)

Multiplying the numerator and denominator by 3 to clear the fraction, we have:

tan(S + T) = (3sqrt(8) - sqrt(3)) / (3 + sqrt(8)*sqrt(3))

Since sqrt(8) = sqrt(4*2) = 2sqrt(2), and sqrt(3) cannot be simplified further, the expression becomes:

tan(S + T) = (6sqrt(2) - sqrt(3)) / (3 + 2sqrt(6))

Thus, the exact value of tan(S + T) is (6sqrt(2) - sqrt(3)) / (3 + 2sqrt(6)).