Respuesta :

[tex]\begin{gathered} \frac{1+\tan^2(u)}{1-\tan^2(u)}=\frac{1+\frac{\sin^2(u)}{\cos^2(u)}}{1-\frac{\sin^2(u)}{\cos^2(u)}} \\ \\ \end{gathered}[/tex]

Fill first blank with:

[tex]\sin ^2(u)[/tex][tex]\frac{1+\frac{\sin^2(u)}{\cos^2(u)}}{1-\frac{\sin^2(u)}{\cos^2(u)}}\times\frac{\cos^2(u)}{\cos^2(u)}=\frac{\frac{\cos ^2(u)+\sin ^2(u)}{\cos ^2(u)}\times\cos ^2(u)}{\frac{\cos ^2(u)-\sin ^2(u)}{\cos ^2(u)}\times\cos ^2(u)}=\frac{\cos ^2(u)+\sin ^2(u)}{\cos ^2(u)-\sin ^2(u)}[/tex]

Fill second blank with:

[tex]\sin ^2(u)[/tex]

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