Question

(1/(cos alpha) + tg*alpha)(1/(cos alpha) - tg*alpha)​

Answer 1

Объяснение:

$\displaystyle\\(\frac{1}{cos\alpha}+tg\alpha })(\frac{1}{cos\alpha } -tg\alpha )=(\frac{1}{cos\alpha } )^2-tg^2\alpha =\frac{1}{cos^2\alpha } -\frac{sin^2\alpha }{cos^2\alpha } =\\\\\\=\frac{1-sin^2\alpha }{cos^2\alpha } =\frac{cos^2\alpha }{cos^2\alpha } =1.$