Bonsoir,
-------------------------------------------------------
Rappels de cours :
∀x∈ℝ, sin²(x)+cos²(x) = 1
∀x∈ℝ\{(π/2)+πℤ}, sec(x) = 1/(cos(x)), donc cos(x)sec(x) = 1
∀x∈ℝ\{πℤ}, csc(x) = 1/(sin(x)), donc sin(x)csc(x) = 1
∀x∈ℝ\{πℤ/2}, sec²(x)+csc²(x) = sec²(x)csc²(x)
-------------------------------------------------------
a) ∀x∈ℝ,
sin(x)(sin(x)+cos(x))-cos(x)(sin(x)-cos(x)) = sin²(x)+sin(x)cos(x)-sin(x)cos(x)+cos²(x) = sin²(x)+cos²(x) = 1
b) ∀x∈ℝ\{πℤ/2},
B = (cos(x)+sec(x))²+(sin(x)+csc(x))²-(sec(x)csc(x))² = cos²(x)+2+sec²(x)+sin²(x)+2+csc²(x)-sec²(x)csc²(x) = cos²(x)+sin²(x)+2+2+sec²(x)+csc²(x)-sec²(x)csc²(x) = 1+2+2+sec²(x)csc²(x)-sec²(x)csc²(x) = 5+sec²(x)csc²(x)-sec²(x)csc²(x) = 5
