Question

Спростіть вираз: , якщо 90°<α<180°.

Answer 1

$\displaystyle\bf\\\sqrt{\frac{1+Sin\alpha }{1-Sin\alpha } } -\sqrt{\frac{1-Sin\alpha }{1+Sin\alpha } } =\frac{\sqrt{1+Sin\alpha } }{\sqrt{1-Sin\alpha } } -\frac{\sqrt{1-Sin\alpha } }{\sqrt{1+Sin\alpha } } =\\\\\\=\frac{\sqrt{1+Sin\alpha } \cdot\sqrt{1+Sin\alpha } -\sqrt{1-Sin\alpha } \cdot\sqrt{1-Sin\alpha } }{\sqrt{1-Sin\alpha } \cdot\sqrt{1+Sin\alpha } } =\\\\\\=\frac{\sqrt{(1+Sin\alpha )(1+Sin\alpha )} -\sqrt{(1-Sin\alpha )(1-Sin\alpha )} }{\sqrt{(1+Sin\alpha )(1-Sin\alpha )} } =$

$\displaystyle\bf\\=\frac{\sqrt{(1+Sin\alpha )^{2} }-\sqrt{(1-Sin\alpha )^{2} } }{\sqrt{1-Sin^{2} \alpha } } =\frac{1+Sin\alpha -(1-Sin\alpha )}{\sqrt{Cos^{2} \alpha } } =\\\\\\=\frac{1+Sin\alpha -1+Sin\alpha }{-Cos\alpha } =-\frac{2Sin\alpha }{Cos\alpha } =-2tg\alpha$