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Log1/8cosec2 π/8 sin2 3π/8 =?

Vikas TU , 13 Years ago
Grade 12th Pass
anser 1 Answers
APURV GOEL

Last Activity: 13 Years ago

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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