Respuesta :

gmany

[tex](1+\sin x)(1-\sin x)=\dfrac{1}{\sec^2x}\\\\L_s=1^2-\sin^2x=1-\sin^2x=\cos^2x\\\\Used:\ a^2-b^2=(a-b)(a+b)\ and\ \sin^2x+\cos^2x=1\to\cos^2x=1-\sin^2x\\\\R_s=\dfrac{1}{\left(\frac{1}{\cos x}\right)^2}=\dfrac{1}{\frac{1}{\cos^2x}}=1\cdot\dfrac{\cos^2x}{1}=\cos^2x\\\\L_s=R_s[/tex]