Question

Решите пожалуйста 25.6

Answer 1

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

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

Answer 2

Ответ:

$1)\ \ tg(a+\beta )-tga\cdot tg\beta \cdot tg(a+\beta )=tg(a+\beta )\cdot (1-tga\cdot tg\beta )=\\\\\\=\Big[\ tg(a+\beta )=\dfrac{tga+tg\beta }{1-tga\cdot tg\beta }\ \ \to \ \ \ 1-tga\cdot tg\beta =\dfrac{tga+tg\beta }{tg(a+\beta )}\ \Big]=\\\\\\=tg(a+\beta )\cdot \dfrac{tga+tg\beta }{tg(a+\beta )}=tga+tg\beta$

$2)\ \ tg(a+\beta )\cdot tg(a-\beta )\cdot (1-tg^2a\cdot tg^2\beta )=\\\\\\=\dfrac{tga+tg\beta }{1-tga\cdot tg\beta }\cdot \dfrac{tga-tg\beta }{1+tga\cdot tg\beta }\cdot (1-tg^2a\cdot tg^2\beta )=\\\\\\=\dfrac{(tga+tg\beta )(tga-tg\beta )}{1-tg^2a\cdot tg^2\beta }\cdot (1-tg^2a\cdot tg^2\beta )=tg^2a-tg^2\beta$