$tan^2x+cot^2x=2sin^5\left (x+\frac{\pi}{4} \right )$
Giải pt $tan^2x+cot^2x=2sin^5\left (x+\frac{\pi}{4} \right )$
Started By vanhongha, 08-09-2012 - 14:46
#1
Posted 08-09-2012 - 14:46
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