Question

Daca x este masura unui unghi ascutit aratati ca: a) cos²x=ctg²x supra 1+ctg²x b) sin²x=tg²x supra 1+tg²x

Answer 1

   
[tex]\displaystyle a) \\ \\ \cos^2x = \frac{\text{ctg}^2 x}{1+\text{ctg}^2 x} = \\ \\ \\ = \frac{ \frac{\cos^2x}{\sin^2x} }{1+\frac{\cos^2x}{\sin^2x} } = \frac{ \frac{\cos^2x}{\sin^2x} }{\frac{\sin^2x}{\sin^2x} +\frac{\cos^2x}{\sin^2x} } = \frac{ \frac{\cos^2x}{\sin^2x} }{\frac{\sin^2x +\cos^2x}{\sin^2x} } = [/tex]



[tex]\displaystyle \\ = \frac{ \frac{\cos^2x}{\sin^2x} }{\frac{1}{\sin^2x} } =\frac{\cos^2x}{\sin^2x} \cdot \frac{\sin^2x}{1 } = \frac{\cos^2x \cdot \sin^2x}{\sin^2x} = \boxed{\cos^2x } \\ \\ cctd[/tex]



$\displaystyle a) \\ \\ \sin^2x = \frac{\text{tg}^2 x}{1+\text{tg}^2 x} = \\ \\ \\ = \frac{ \frac{\sin^2x}{\cos^2x} }{1+\frac{\sin^2x}{\cos^2x} } = \frac{ \frac{\sin^2x}{\cos^2x} }{\frac{\cos^2x}{\cos^2x} +\frac{\sin^2x}{\cos^2x} } = \frac{ \frac{\sin^2x}{\cos^2x} }{\frac{\cos^2x +\sin^2x}{\cos^2x} } =$


[tex]\displaystyle \\ = \frac{ \frac{\sin^2x}{\cos^2x} }{\frac{1}{\cos^2x} } = \frac{\sin^2x}{\cos^2x} \cdot \frac{\cos^2x}{1 } = \frac{\sin^2x \cdot \cos^2x}{\cos^2x} = \boxed{\sin^2x } \\ \\ cctd[/tex]