Question

Найдите tg(α), если sin=

Answer 1

$sina=\frac{3}{\sqrt{10}}>0\; \; \Rightarrow \; \; a\in (2\pi n\,;\pi +2\pi n)\\\\\boxed{sin^2a+cos^2a=1}\; \; \to \; \; cos^2a=1-sin^2a=1-\frac{9}{10}=\frac{1}{10}\; ,\; \; cosa=\pm \frac{1}{\sqrt{10}}\\\\\boxed {tga=\frac{sina}{cosa}}\; \; ,\; \; tga=\frac{\frac{3}{\sqrt{10}}}{\pm \frac{1}{\sqrt{10}}}=\pm \frac{3}{1}=\pm 3\\\\\\Esli\; \; a\in (2\pi n\, ;\frac{\pi }{2}+2\pi n)\; ,\; to\; \; tga=3\; .\\\\Esli\; a\in (\frac{\pi}{2}+2\pi n\, ;\pi +2\pi n)\; \; ,\; to \; \; tga=-3\; .$

Answer 2

It seems like there is a typo in your question. The expression "sin=" is incomplete, and the value of the sine function is missing. To find the tangent of an angle (α), you need the values of both the sine and cosine functions, which are related by the identity:

$\tan(\alpha) = \frac{\sin(\alpha)}{\cos(\alpha)}$

If you provide the value of the sine function (sin(α)) or the cosine function (cos(α)), I can help you find the tangent (tg(α)). Please provide the missing information, and I'll be glad to assist you further!