Respuesta :

mathmate
cos⁡(x)/(1+sin⁡(x) )+(1+sin⁡(x))/cos⁡(x) =2sec⁡(x)

Work on the left hand side.
[Common denominator is (1+sin(x))*cos(x)]
cos⁡(x)/(1+sin⁡(x) )+(1+sin⁡(x))/cos⁡(x)    
= (cos(x)^2+(1+sin(x))^2)/((1+sin(x))*cos(x))
=(cos(x)^2+1+sin(x)^2+2sin(x))/((1+sin(x))*cos(x))
=(cos(x)^2+sin(x)^2+1+2sin(x))/((1+sin(x))*cos(x))
=(1+1+2sin(x))/((1+sin(x))*cos(x))
=(2+2sin(x))/((1+sin(x))*cos(x))
=2(1+sin(x))/((1+sin(x))*cos(x))
=2/cos(x)
=2 sec(x)                [QED]
gmany
[tex]\dfrac{\cos x}{1+\sin x}+\dfrac{1+\sin x}{\cos x}=2\sec x\\\\L_s=\dfrac{\cos x\cos x}{\cos x(1+\sin x)}+\dfrac{(1+\sin x)(1+\sin x)}{\cos x(1+\sin x)}\\\\=\dfrac{\cos^2x+1+\sin x+\sin x+\sin^2x}{\cos x(1+\sin x)}=\dfrac{(\cos^2x+\sin^2x)+1+2\sin x}{\cos x(1+\sin x)}\\\\=\dfrac{1+1+2\sin x}{\cos x(1+\sin x)}=\dfrac{2+2\sin x}{\cos x(1+\sin x)}=\dfrac{2(1+\sin x)}{\cos x(1+\sin x)}\\\\=\dfrac{2}{\cos x}=2\cdot\dfrac{1}{\cos x}=2\sec x=R_s[/tex]

[tex]\text{Used:}\\\\\sin^2\alpha+\cos^2\alpha=1\\\\\sec\alpha=\dfrac{1}{\cos\alpha}[/tex]

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