[tex] \frac{1+ctg2 \alpha }{1-ctg2 \alpha } = \frac{sin(2 \alpha- \frac{\pi }{4} ) }{sin(2 \alpha- \frac{\pi }{4} ) } [/tex]. Created by масяня3009. algebra-ru

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[tex] \frac{1+ctg2 \alpha }{1-ctg2 \alpha } = \frac{sin(2 \alpha- \frac{\pi }{4} ) }{sin(2 \alpha- \frac{\pi }{4} ) } [/tex]

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Artem112

Verified answer

(1+ctg2a) / (1 -ctg2a) = sin (2a -п/4) /sin (2a -п/4) = 1
(1+ctg2a) = (1 -ctg2a)
2*(ctg2a) = 0
ctg2a = 0
2a = п/2 + пk , k Є Z
a = п/4 + пk/2 , k Є Z
ОТВЕТ a = п/4 + пk/2 , k Є Z

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