Question

найти tg & и ctg & если a)cos&=1/5 b)sin&=2/3

Answer 1

Объяснение:

а).

$1 + {tg}^{2} \beta = \frac{1}{ {cos}^{2} \beta } \\ cos \beta = \frac{1}{5} \\ 1 + {tg}^{2} \beta = 1 \div {( \frac{1}{5}) }^{2} \\ 1 + {tg}^{2} \beta = 25 \\ {tg}^{2} \beta = 24 \\ tg \beta = \sqrt{24} \\ \sqrt{24 } = 2 \sqrt{6}$

$tg \beta \times ctg \beta = 1 \\ ctg \beta = \frac{1}{tg \beta } \\ ctg \beta = \frac{1}{ \sqrt{24} } \\ \frac{1}{ \sqrt{24} } = \frac{ \sqrt{24} }{24} = \frac{2 \sqrt{6} }{24} = \frac{ \sqrt{6} }{12} \\ ctg \beta = \frac{ \sqrt{6} }{12}$

b).

$1 + {ctg}^{2} \beta = \frac{1}{ {sin}^{2} \beta } \\ sin \beta = \frac{2}{3} \\ 1 + {ctg}^{2} \beta = 1 \div {( \frac{2}{3})}^{2} \\ 1 + {ctg}^{2} \beta = \frac{9}{4} \\ {ctg}^{2} \beta = \frac{5}{4} \\ ctg \beta = \sqrt{ \frac{5}{4} } \\ tg \beta = \frac{1}{ctg \beta } \\ tg \beta = \sqrt{ \frac{4}{5} } \\ tg \beta = \frac{2}{ \sqrt{5} }$