Question

(tg(a-b)-tga+tgb)/(tg(a-b)*tgb)

Answer 1

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

$=\frac{tg\alpha\cdot tg^2\beta-tg^2\alpha\cdot tg\beta}{(tg\alpha-tg\beta)\cdot tg\beta}=\frac{-tg\alpha\cdot tg\beta \cdot(tg\alpha -tg\beta) }{(tg\alpha-tg\beta)\cdottg\beta } = -tg\alpha$

Answer 2

To simplify the expression $\frac{{\tan(a-b) - \tan(a) + \tan(b)}}{{\tan(a-b) \cdot \tan(b)}}$, we can start by factoring out $\tan(a-b)$ from the numerator:

$$\frac{{\tan(a-b) - \tan(a) + \tan(b)}}{{\tan(a-b) \cdot \tan(b)}} = \frac{{\tan(a-b) \cdot (1 - \frac{{\tan(a)}}{{\tan(a-b)}} + \frac{{\tan(b)}}{{\tan(a-b)}})}}{{\tan(a-b) \cdot \tan(b)}}$$

Next, we can simplify the expression by canceling out the common factor $\tan(a-b)$ from the numerator and denominator:

$$\frac{{\cancel{\tan(a-b)} \cdot (1 - \frac{{\tan(a)}}{{\cancel{\tan(a-b)}}}} + \frac{{\tan(b)}}{{\cancel{\tan(a-b)}})}}{{\cancel{\tan(a-b)}} \cdot \tan(b)} = \frac{{1 - \frac{{\tan(a)}}{{\tan(a-b)}} + \frac{{\tan(b)}}{{\tan(a-b)}}}}{{\tan(b)}}$$

Thus, the simplified expression is $\frac{{1 - \frac{{\tan(a)}}{{\tan(a-b)}} + \frac{{\tan(b)}}{{\tan(a-b)}}}}{{\tan(b)}}$.