Question

tga/1+tga+tga/1-tga=tg2a Доведіть тотожність

Answer 1

Ответ:

$\frac{tg\alpha}{1+tg\alpha}+\frac{tg\alpha}{1-tg\alpha}=tg2\alpha$

Объяснение:

$\frac{tg\alpha}{1+tg\alpha}+\frac{tg\alpha}{1-tg\alpha}=\frac{tg\alpha(1-tg\alpha)}{(1+tg\alpha)(1-tg\alpha)}+\frac{tg\alpha(1+tg\alpha)}{(1+tg\alpha)(1-tg\alpha)}=$

$\frac{tg\alpha(1-tg\alpha)+tg\alpha(1+tg\alpha)}{1-tg^2\alpha}=\frac{tg\alpha-tg^2\alpha+tg\alpha+tg^2\alpha}{1-tg^2\alpha}=$

$\frac{2tg\alpha}{1-tg^2\alpha}=\frac{2\cdot\frac{sin\alpha}{cos\alpha}}{1-\frac{sin^2\alpha}{cos^2\alpha}}=\frac{2tg\alpha}{1-tg^2\alpha}=\frac{\frac{2sin\alpha}{cos\alpha}}{\frac{cos^2\alpha}{cos^2\alpha}-\frac{sin^2\alpha}{cos^2\alpha}}=$

$\frac{\frac{2sin\alpha}{cos\alpha}}{\frac{cos^2\alpha}{cos^2\alpha}-\frac{sin^2\alpha}{cos^2\alpha}}=\frac{\frac{2sin\alpha}{cos\alpha}}{\frac{cos^2\alpha-sin^2\alpha}{cos^2\alpha}}=$

$\frac{2sin\alpha}{cos\alpha}\cdot\frac{cos^2\alpha}{cos^2\alpha-sin^2\alpha}=\frac{2sin\alpha cos\alpha}{cos^2\alpha-sin^2\alpha}=\frac{sin 2\alpha}{cos2\alpha}=tg2\alpha$