Question

(I-tga)^2+(l+tga)^2=2/cos^2a

Answer 1

$(1-tga)^2+(1+tga)^2=1-2tga+tg^2a+1+2tga+tg^2a=\\=2(1+tg^2a)=2cdot frac{1}{cos^2a}=frac{2}{cos^2a}\\\Osnovnoe; trigon.; tozdestvo:; 1+tg^2a=frac{1}{cos^2a}.\\1+tg^2a=1+frac{sin^2a}{cos^2a}=frac{cos^2a+sin^2a}{cos^2a}=frac{1}{cos^2a}$