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Log1/8cosec2 π/8 sin2 3π/8 =?
Log1/8cosec2 π/8 sin2 3π/8 = Log 1/(2√2 sinΠ/8) sin3Π/8
= Log 1/(2√2 sinΠ/8) cosΠ/8
we have sinΠ/4 = 2.sinΠ/8.cosΠ/8
so sinΠ/8 = (sinΠ/4)/2.cosΠ/8
so 1/(2√2 sinΠ/8) = cosΠ/8
hence Log 1/(2√2 sinΠ/8) cosΠ/8 = LogcosΠ/8cosΠ/8 = 1
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