If a and b are angles lying in the 2nd quadrant such that cosa= -2/3 and cscb=2 find the following sin(a+b) and tan(a+b)

Both a and b lie in the second quadrant, which means

π/2 < a < π

π/2 < b < π

for which we have both cos(a) and cos(b) negative, while both sin(a) and sin(b) are positive. We're also given that

cos(a) = -2/3

csc(b) = 2   =>   sin(b) = 1/2

Now, recall the following identities:

• sin^2(x) + cos^2(x) = 1

• sin(x + y) = sin(x) cos(y) + cos(x) sin(y)

• cos(x + y) = cos(x) cos(y) - sin(x) sin(y)

• tan(x) = sin(x)/cos(x)

Now,

sin(a + b) = sin(a) cos(b) + cos(a) sin(b)

… = √(1 - cos^2(a)) (-√(1 - sin^2(b))) + cos(a) sin(b)

* * where we take the negative square root because we expect cos(b) < 0 * *

… = - √(1 - (-2/3)^2) √(1 - (1/2)^2) + (-2/3) (1/2)

… = - √(1 - 4/9) √(1 - 1/4) - (2/3) (1/2)

… = - √(5/9) √(3/4) - 1/3

… = - (√5/3) (√3/2) - 1/3

… = -√15/6 - 1/3

… = -(√15 + 2) / 6

From the above, we know sin(a) = √5/3 and cos(b) = -√3/2. Then

cos(a + b) = cos(a) cos(b) - sin(a) sin(b)

… = (-2/3) (-√3/2) - (√5/3) (1/2)

… = √3/3 - √5/6

… = (2√3 - √5) / 6

and so it follows that

tan(a + b) = sin(a + b) / cos(a + b)

… = (-(√15 + 2) / 6) / ((2√3 - √5) / 6)

… = - (√15 + 2) / (2√3 - √5)

… = - (√15 + 2) (2√3 + √5) / ((2√3 - √5) (2√3 + √5))

… = - (2√45 + 4√3 + √75 + 2√5) / (12 - 5)

… = - (6√5 + 4√3 + 5√3 + 2√5) / 7

… = - (8√5 + 9√3) / 7

stops on an even number.

range is infinite on both x and y axis positive and negative

step-by-step explanation: