1) sin²α/cos²α +tqα *ctqα=tq²α +1= 1/cos²α ;
2) т.к. α ∈ (3/2π ;2π) то
sinα = - sqrt(1-cos²α) = - sqrt(1 - 0,6²) = -sqrt(1-0,36)= - sqrt0,64 = -0,8 ;
tqα = sinα/cosα = - 4/3.
3)
а)2cos0° +5cos90° - 4tq180° = 2+5 -0=7;
б)ctqπ/2+3cosπ/2 - 4sin3π/2= 0+0 - 4*(-1)= 4 ;
в) tq45°cos30°ctq60° = 1*√3/2*√3/3 = 1/2 .
4)
а)sin(π/2 +α)+cos(π + α)+ctq(2π -α)= cosα - cosα -ctqα = -ctqα;
[ формулы приводения ]
б) cos(α +π/2)cos(3π - α)+sin(α +5π/2) = -sinα*(cos(2π+π- α)+sin(2π+π/2+α)=
= -sinα*cos(π-α)+sin(π/2+α) = -sinα*(-cosα)+cosα = cosα(1+sinα).