Respuesta :
Answer:
Proof is in the explanation.
Step-by-step explanation:
I figure out what you were saying:
[tex]\frac{\sec(x)-1}{\sec(x)+1}=\frac{\sin^2(x)}{(1+\cos(x))^2}[/tex]
I'm going to try to rewrite left hand side as right hand side.
I'm going to rewrite [tex]\sec(x)=\frac{1}{\cos(x)}[/tex]:
[tex]\frac{\frac{1}{\cos(x)}-1}{\frac{1}{\cos(x)}+1}[/tex]
Multiply both numerator and denominator by [tex]\cos(x)[/tex]:
[tex]\frac{1-\cos(x)}{1+\cos(x)}[/tex]
Multiply both numerator and denominator by [tex]1+\cos(x)[/tex]:
[tex]\frac{(1-\cos(x))(1+\cos(x))}{(1+\cos(x))^2}[/tex]
When multiplying conjugates you only have to do first time first and last times last:
[tex]\frac{1-\cos^2(x)}{(1+\cos(x))^2}[/tex]
We can rewrite the numerator using the Pythagorean Identity: [tex]\cos^2(x)+\sin^2(x)=1[/tex].
This gives us:
[tex]\frac{\sin^2(x)}{(1+\cos(x))^2}[/tex]
