Question

2.17) prove que:

(tgx+1/cosx)²=1+senx/1-senx​

Answer 1

$(tga+\frac{1}{cosa})^2 = (tga+seca)^2=tg^2a+2*tga*seca+sec^2a =\\\\= (\frac{sena}{cosa})^2+2*\frac{sena}{cosa}*\frac{1}{cosa}+\frac{1}{(cosa)^2} = \frac{sen^2a}{cos^2a}+\frac{2sena}{cos^2a}+\frac{1}{cos^2a}=\frac{sen^2a+2sena+1}{cos^2a}=\frac{(sena+1)^2}{1-sen^2a}=\frac{(sena+1)(sena+1)}{(1+sena)(1-sena)}=\frac{1+sena}{1-sena}$