Respuesta :

konrad509
[tex]\dfrac{1+\sin x}{\cos x}=\dfrac{\cos (-x)}{1+\sin(-x)}\\ \dfrac{1+\sin x}{\cos x}=\dfrac{\cos x}{1-\sin x}\\ \cos^2 x=(1+\sin x)(1-\sin x)\\ \cos^2 x=1-\sin^2 x\\ \sin^2 x+\cos^2 x=1 [/tex]