Question

,
6tg^2 a, если sin a=1/корень из 3

Answer 1

$Sin\alpha=\dfrac{1}{\sqrt{3} }\\\\1+Ctg^{2}\alpha=\dfrac{1}{Sin^{2} \alpha } \\\\Ctg^{2}\alpha= \dfrac{1}{Sin^{2} \alpha }-1=\dfrac{1}{\Big(\dfrac{1}{\sqrt{3}} \Big)^{2} }-1=\dfrac{1}{\dfrac{1}{3} }-1= 3-1=2\\\\tg^{2}\alpha =\dfrac{1}{Ctg^{2}\alpha}=\frac{1}{2}\\\\6tg^{2}\alpha =6\cdot\dfrac{1}{2}=3\\\\\boxed{6tg^{2}\alpha =3}$

Answer 2

$\sin( \alpha ) = \frac{1}{ \sqrt{3} } \\ \\ \cos {}^{2} ( \alpha ) = 1 - \sin {}^{2} ( \alpha ) = 1 - \frac{1}{3} = \frac{2}{3}$

Далее используем формулу:

$1 + {tg}^{2} \alpha = \frac{1}{ \cos {}^{2} ( \alpha ) } \\ tg {}^{2} \alpha = \frac{1}{ \cos {}^{2} ( \alpha ) } - 1 = \frac{3}{2} - 1 = \frac{1}{2} = 0.5 \\ \\ 6 {tg}^{2} \alpha = 6 \times 0.5 = 3$

ответ: 3