Доказать тождество:

 

[tex]\frac{(Sin t + Cos t)^2 - 1}{Ctg t - Sin t Cos t} = 2 Tg^2 t[/tex]

(sin t +cos t)^2 - 1/ (ctg t - sin t*cos t) = =(sin t)^2 +2sin t*cos t+(cos t)^2 - ((sin t)^2+(cos t)^2)/ (ctg t - sin t*cos t) = 

=(sin t)^2 +2sin t*cos t+(cos t)^2 - (sin t)^2-(cos t)^2/ (ctg t - sin t*cos t) = =2sin t*cos t/ (cos t/sin t) - sin t*cos t = 2sin t*cos t/ (cos t - (sin t)^2*cos t)/sin t =  = 2sin t*cos t/ (cos t (1- (sin t)^2)/sin t == 2sin t/  (1- (sin t)^2)/sin t = = 2(sin t)^2/  (1- (sin t)^2) = =2(sin t)^2/  (1- (sin t)^2)== 2(sin t)^2/  ((sin t)^2+(cos t)^2- (sin t)^2)= =2(sin t)^2/   (cos t)^2 = 2 (tg t)^2