Respuesta :

Answer:

cos x ≠ 0 ⇔ x ≠ [tex]\frac{pi}{2}+k.pi[/tex] ; k ∈ N

[tex](1-cosx)(1+\frac{1}{cosx})=sinx.tanx\\\\<=>(1-cosx)(\frac{cosx+1}{cosx})=sinx.\frac{sinx}{cosx}\\\\<=> \frac{1-cos^{2} x}{cosx} =\frac{sin^{2}x }{cosx}\\\\=>1-cos^{2}x=sin^{2}x[/tex]

<=> cos²x + sin²x = 1

⇔ 1 = 1

=> x =  { R \ (pi/2 + k.pi); k ∈ N}

Step-by-step explanation: