Question

Помогите найти tg(a-π/6) ,если ctg a =1/√2

Answer 1

$tg( \alpha - \frac{ \pi }{6})= \frac{tg \alpha -tg \frac{ \pi }{6} }{1+tg \alpha tg \frac{ \pi }{6} } \\\\tg \alpha = \frac{1}{Ctg \alpha } = \frac{1}{ \frac{1}{ \sqrt{2} } } = \sqrt{2} \\\\tg \frac{ \pi }{6}= \frac{ \sqrt{3} }{3} \\\\\\tg( \alpha - \frac{ \pi }{6})= \frac{ \sqrt{2} - \frac{ \sqrt{3} }{3} }{1+ \sqrt{2}* \frac{ \sqrt{3} }{3} } = \frac{ \frac{3 \sqrt{2}- \sqrt{3} }{3} }{1+ \frac{ \sqrt{6} }{3} } = \frac{3 \sqrt{2}- \sqrt{3} }{3+ \sqrt{6} } =$$= \frac{ \sqrt{3}( \sqrt{6}-1) }{ \sqrt{3}( \sqrt{3} - \sqrt{2}) } = \frac{ \sqrt{6}-1 }{ \sqrt{3}- \sqrt{2} }$