S'il vous plaît aidez moi c'est un devoir maison

Bonsoir,

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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)
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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