Answer:
Proved
Step-by-step explanation:
Required
Show that:
[tex]\frac{1+cos2a+sin2a}{1-cos2a+sin2a}=cota[/tex]
[tex]cos2a = cos(a + a) = cos^2a - sin^2a[/tex]
So, we have:
[tex]\frac{1+cos^2a - sin^2a+sin2a}{1-(cos^2a - sin^2a)+sin2a}=cota[/tex]
[tex]\frac{1+cos^2a - sin^2a+sin2a}{1-cos^2a + sin^2a+sin2a}=cota[/tex]
[tex]1- cos^2a = sin^2a[/tex]
So, we have:
[tex]\frac{1+cos^2a - sin^2a+sin2a}{ sin^2a+ sin^2a+sin2a}=cota[/tex]
[tex]\frac{1+cos^2a - sin^2a+sin2a}{2sin^2a+sin2a}=cota[/tex]
Rearrange the numerator
[tex]\frac{1 - sin^2a+cos^2a+sin2a}{2sin^2a+sin2a}=cota[/tex]
[tex]1- sin^2a= cos^2a[/tex]
So, we have
[tex]\frac{cos^2a+cos^2a+sin2a}{2sin^2a+sin2a}=cota[/tex]
[tex]\frac{2cos^2a+sin2a}{2sin^2a+sin2a}=cota[/tex]
[tex]sin2a = 2sina\ cosa[/tex]
So, we have:
[tex]\frac{2cos^2a+2sinacosa}{2sin^2a+2sinacosa}=cota[/tex]
Factorize:
[tex]\frac{cosa(2cosa+2sina)}{sina(2sina+2cosa)}=cota[/tex]
Rewrite as:
[tex]\frac{cosa(2cosa+2sina)}{sina(2cosa+2sina)}=cota[/tex]
[tex]\frac{cosa}{sina} = cota[/tex]
[tex]cota = cota[/tex]
