Respuestas

2013-11-02T20:21:23+01:00

pregunta a
 \frac{\sec  \alpha +\csc  \alpha }{\sec  \alpha -\csc  \alpha } = \frac{\sin  \alpha +\cos  \alpha  }{\sin  \alpha -\cos  \alpha }

 \frac{ \frac{1}{\cos  \alpha }+  \frac{1}{\sin  \alpha } }{ \frac{1}{\cos  \alpha } - \frac{1}{\sin  \alpha } }=\frac{\sin \alpha +\cos \alpha }{\sin \alpha -\cos \alpha }

 \frac{ \frac{\sin \alpha +\cos \alpha}{\sin  \alpha *\cos  \alpha } }{ \frac{\sin \alpha -\cos \alpha}{\sin  \alpha *\cos  \alpha } } =\frac{\sin \alpha +\cos \alpha }{\sin \alpha -\cos \alpha }

se cancela arriba y abajo \sin  \alpha *\cos  \alpha
y me queda la igualdad


\frac{\sin \alpha +\cos \alpha }{\sin \alpha -\cos \alpha }=\frac{\sin \alpha +\cos \alpha }{\sin \alpha -\cos \alpha }


pregunta b
 (acosx+bsenx)^{2} + (asenx-bcosx)^{2} = a^{2}+ b^{2}

a^{2}cos^{2}x+b^{2}sen^{2}x+a^{2}sen^{2}x+b^{2}cos^{2}x= a^{2} + b^{2}

a^{2}(sen^{2}x+cos^{2}x)+b^{2}(sen^{2}x+cos^{2}x)= a^{2}+  b^{2}

pero (sen^{2}x+cos^{2}x)=1

entonces os queda 

 (a^{2}+  b^{2})(1)=  a^{2}  +b^{2}