1-4sin^2t*cos^2t/(sint+cost)^2+2sint*cost=1

Во  всей вероятности  (1- 4sin²t·cos²t)/(sint+cost)² + 2sint·cost) =1  (1- sin2t)²/(1+2sint·cost) +sin2t = 1   1- sin2t + sin2t + (sin2t)² = 1+sin2t    ;    (sin2t)² - sin2t =0   sin2t·(sin2t -1) = 0    a)   sin2t =0    ⇒  2t = πk  ⇒                 t= π/2 ·k    ;  k∈Z    b)   sin2t =1    ⇒  2t = 2πm + π/2   ⇒    t=πm +π/4 ; m∈Z