Answer:
a) cos(B) = -2√(2k)/(k+2)
b) k > 2
Explanation:
Given csc(B) = (k+2)/(k-2), where B is in the second quadrant, you want cos(B) and the restrictions on k.
Relations
The trig relations needed to solve this are ...
- csc(x) = 1/sin(x)
- sin²(x) +cos²(x) = 1
- sin(x) > 0, cos(x) < 0 in the 2nd quadrant
a) Cosine
The cosine can be found from ...
sin(B) = 1/csc(B) = (k -2)/(k +2)
cos(B) = √(1 -sin²(B))
cos(B) = -√(1 -((k -2)/(k +2))²) = -√((k² +4k +4) -(k² -4k +4))/(k+2)
cos(B) = -2√(2k)/(k+2) . . . . . . . sign is negative in 2nd quadrant
b) Restrictions
In order for csc((k+2)/(k-2)) to be defined, we must have ...
- k ≠ 2 . . . . . . . . . . . the denominator cannot be zero
- (k+2)/(k-2) ≥ 1 . . . . cosecant can only have values of 1 or greater
We can rewrite the second restriction as ...
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Additional comment
We find it convenient to rewrite the restriction inequality with 0 on one side so we don't ever have to multiply by (k-2). That would generated two inequalities, one for k<2 and one for k>2. That seems like more work. (The solution that way is the same.)
We note that 4/(k-2) can never be zero. Conveniently, k>2 also avoids k = 2, which makes the denominator zero.
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