Ответ:
[tex]tga=\frac{1}{5}\\\\26cos^2a-1=26\cdot \frac{1+cos2a}{2}-1=13\cdot (1+cos2a)-1=13\cdot cos2a+12=\\\\=13\cdot \frac{1-tg^2a}{1+tg^2a}+12=13\cdot \frac{1-\frac{1}{25}}{1+\frac{1}{25}}+12=13\cdot \frac{24}{26}+12=\frac{24}{2}+12=12+12=24\\\\\\Formylu:\\\\\star \; \; cos^2x=\frac{1+cos2x}{2}\; \; \star \\\\\star \; \; cos2x=cos^2x-sin^2x=\frac{cos^2x-sin^2x}{cos^2x+sin^2x}=[\, \frac{:cos^2x}{:cos^2x}\, ]=\frac{1-tg^2x}{1+tg^2x}\; \; \star[/tex]