Asked by Anonymous
If 0<x<1 is such that cscx - secx = (13^1/2)/6, then cotx - tanx equals
Answers
Answered by
m
correction: 0<x<pi/4
Answered by
oobleck
cosx - sinx = √13/6 sinx cosx
6cosx - 6sinx = √13 sinx cosx
36cos^2x - 72 sinx cosx + 36sin^2x = 13 sin^2x cos^2x
13 sin^2x cos^2x + 72 sinx cosx - 36 = 0
13 sin^2 2x + 144 sin2x - 144 = 0
sin2x = 12/13
So now we have
cotx - tanx = cosx/sinx - sinx/cosx
= (cos^2x - sin^2x)/(sinx cosx) = 2cos2x/sin2x
= 2√(1 - (12/13)^2)/(12/13)
= 5/6
huh -- better double-check my math; seems too simple
6cosx - 6sinx = √13 sinx cosx
36cos^2x - 72 sinx cosx + 36sin^2x = 13 sin^2x cos^2x
13 sin^2x cos^2x + 72 sinx cosx - 36 = 0
13 sin^2 2x + 144 sin2x - 144 = 0
sin2x = 12/13
So now we have
cotx - tanx = cosx/sinx - sinx/cosx
= (cos^2x - sin^2x)/(sinx cosx) = 2cos2x/sin2x
= 2√(1 - (12/13)^2)/(12/13)
= 5/6
huh -- better double-check my math; seems too simple
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