Question

(1+tga/1-tga)-(1+sin2a/cos2a)

Answer 1

$\frac{1+tga}{1-tga} - \frac{1+sin2a}{cos2a}= \frac{1+tga}{1-tga}- \frac{1+ \frac{2tga}{1+tg^2a} }{ \frac{1-tg^2a}{1+tg^2a} }= \frac{1+tga}{1-tga}- \frac{ \frac{1+tg^2a+2tga}{1+tg^2a} }{ \frac{1-tg^2a}{1+tg^2a} }=\\\\= \frac{1+tga}{1-tga}- \frac{1+t^2a+2tga}{1-tg^2a}= \frac{(1+tga)^2-(1+tg^2a+2tga)}{1-tg^2a}=\\\\= \frac{1+2tga+tg^2a-1-tg^2a-2tga}{1-tg^2a}= \frac{0}{1-tg^2a}=0$