Question

(tgx+ctgx)*(1-cos4x)=4sin2x

Answer 1

$(tgx+ctgx)(1-cos4x)=4sin2x\ (frac{sinx}{cosx}+frac{cosx}{sinx})( 8cos^2x-8cos^4x) =\ frac{sinx}{cosx}*(8cos^2x-8cos^4x)+frac{cosx}{sinx}*(8cos^2x-8cos^4x)=\ sinx(8cosx-8cos^3x)+8cos^3x*sinx=\ 8cosx*sinx-8cos^3xsinx+8cos^3x*sinx=\ 8cosx*sinx=4*2sinx*cosx=4*sin2x$
 Доказано !