№2 Для решения будем использовать формулу [tex]\large\boxed{ \boxed{\cos 2a = \cos ^2a -\sin^2a}}[/tex][tex]\large \boldsymbol{}\dfrac{\cos \beta }{\cos \frac{\beta }{2}-\sin \frac{\beta }{2} } = \dfrac{\cos^2 \tfrac{\beta }{2}-\sin^2 \frac{\beta }{2}}{\cos \frac{\beta }{2}-\sin \frac{\beta }{2}} = \\\\\\\\ \dfrac{ (\cos \frac{\beta }{2}+\sin \frac{\beta }{2}) (\cos \frac{\beta }{2}-\sin \frac{\beta }{2})}{\cos \frac{\beta }{2}-\sin \frac{\beta }{2}} = \boxed{\cos \tfrac{\beta }{2}+\sin \tfrac{\beta }{2}}[/tex]№4 В этом случае используем формулы[tex]\large \boldsymbol{} 1) ~\sin 2 a = 2\sin a \cos a \\\\ 2) ~ \sin ^2 a + \cos ^2 a= 1[/tex][tex]\large \boldsymbol{} \displaystyle \frac{(\sin a+ \cos a)^2}{1-\sin^2 2a } = \frac{\sin^2a + \cos ^2 a + 2\sin a \cos a }{1-\sin^2 2a} =\\\\\\\ \frac{1+\sin 2a}{(1-\sin 2a)(1+ \sin 2a)} = \boxed{\frac{1}{1-\sin 2a } }[/tex]
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№2
Для решения будем использовать формулу
[tex]\large\boxed{ \boxed{\cos 2a = \cos ^2a -\sin^2a}}[/tex]
[tex]\large \boldsymbol{}\dfrac{\cos \beta }{\cos \frac{\beta }{2}-\sin \frac{\beta }{2} } = \dfrac{\cos^2 \tfrac{\beta }{2}-\sin^2 \frac{\beta }{2}}{\cos \frac{\beta }{2}-\sin \frac{\beta }{2}} = \\\\\\\\ \dfrac{ (\cos \frac{\beta }{2}+\sin \frac{\beta }{2}) (\cos \frac{\beta }{2}-\sin \frac{\beta }{2})}{\cos \frac{\beta }{2}-\sin \frac{\beta }{2}} = \boxed{\cos \tfrac{\beta }{2}+\sin \tfrac{\beta }{2}}[/tex]
№4
В этом случае используем формулы
[tex]\large \boldsymbol{} 1) ~\sin 2 a = 2\sin a \cos a \\\\ 2) ~ \sin ^2 a + \cos ^2 a= 1[/tex]
[tex]\large \boldsymbol{} \displaystyle \frac{(\sin a+ \cos a)^2}{1-\sin^2 2a } = \frac{\sin^2a + \cos ^2 a + 2\sin a \cos a }{1-\sin^2 2a} =\\\\\\\ \frac{1+\sin 2a}{(1-\sin 2a)(1+ \sin 2a)} = \boxed{\frac{1}{1-\sin 2a } }[/tex]