Given that cos(2a) = 15/17 and 180° < 2a < 270°.
We know that cos(2a) = 2cos^2(a) - 1.
Substitute cos(2a) = 15/17 into the equation, we get:
2cos^2(a) - 1 = 15/17
2cos^2(a) = 15/17 + 1
2cos^2(a) = 32/17
cos^2(a) = 16/17
cos(a) = sqrt(16/17) = 4/√17
Since 180° < 2a < 270°, a falls in the third quadrant.In the third quadrant, cos(a) < 0 and tan(a) = -sin(a)/cos(a).
Therefore, cos(a) = -4/√17 and ctg(a) = √17/4.
Given that cos(2a) = 15/17 and 180° < 2a < 270°.
We know that cos(2a) = 2cos^2(a) - 1.
Substitute cos(2a) = 15/17 into the equation, we get:
2cos^2(a) - 1 = 15/17
2cos^2(a) = 15/17 + 1
2cos^2(a) = 32/17
cos^2(a) = 16/17
cos(a) = sqrt(16/17) = 4/√17
Since 180° < 2a < 270°, a falls in the third quadrant.
In the third quadrant, cos(a) < 0 and tan(a) = -sin(a)/cos(a).
Therefore, cos(a) = -4/√17 and ctg(a) = √17/4.