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Ответ:

[tex]- \frac{ \sin(2x) + 1}{2 \cos(2x) } \times \tan( \frac{\pi}{4} - x) =\\=-\frac{2sinxcosx + 1}{2(cos^{2}x-sin^{2}x) } * \tan( \frac{\pi}{4} - x) = -\frac{1}{2} *\frac{cos^{2}x+sin^{2}x + 2sinxcosx}{(cosx-sinx)(cosx+sinx)} * \tan( \frac{\pi}{4} - x) = -\frac{1}{2} * \frac{(sinx+cosx)^{2} }{(cosx-sinx)(cosx+sinx)}* \tan( \frac{\pi}{4} - x) = -\frac{1}{2} * \frac{sinx+cosx}{cosx-sinx} * \tan( \frac{\pi}{4} - x) = -\frac{1}{2}*\frac{tgx+1}{1-tgx} * \frac{tg\frac{\pi }{4}-tgx }{1+tgx*tg\frac{\pi }{4} } =[/tex]

[tex]=-\frac{1}{2}*\frac{tgx+1}{1-tgx} * \frac{1-tgx }{1+tgx } = -\frac{1}{2} = -0.5[/tex]