Answer:
[tex]\frac{1 + sin(2 a)}{cos(2 a) } = \frac{cos(a)+ sin(a)}{cos(a) - sin(a) }[/tex]
Step-by-step explanation:
The question is incomplete; however, I'll simplify the given expression as far as it can be simplified.
Given
[tex]\frac{1 + sin(2 a)}{cos(2 a) }[/tex]
Required
Simplify
[tex]\frac{1 + sin(2 a)}{cos(2 a) }[/tex]
In trigonometry:
[tex]sin(2a) = 2sin(a)cos(a)[/tex]
So, the expression becomes:
[tex]\frac{1 + 2sin(a)(cos(a)}{cos(2 a) }[/tex]
Also in trigonometry:
[tex]cos^2(a) + sin^2(a) = 1[/tex]
So, the expression becomes:
[tex]\frac{cos^2(a) + sin^2(a) + 2sin(a)(cos(a)}{cos(2 a) }[/tex]
Also:
[tex]cos(2a) = cos^2(a) - sin^2(a)[/tex]
So, we have:
[tex]\frac{cos^2(a) + sin^2(a) + 2sin(a)(cos(a)}{cos^2(a) - sin^2(a) }[/tex]
Rearrange the numerator:
[tex]\frac{cos^2(a) + 2sin(a)cos(a) + sin^2(a)}{cos^2(a) - sin^2(a) }[/tex]
Expand the numerator
[tex]\frac{cos^2(a) + sin(a)cos(a)+ sin(a)cos(a) + sin^2(a)}{cos^2(a) - sin^2(a) }[/tex]
Factorize:
[tex]\frac{cos(a)(cos(a) + sin(a))+sin(a)(cos(a)+ sin(a))}{cos^2(a) - sin^2(a) }[/tex]
[tex]\frac{(cos(a) + sin(a))(cos(a)+ sin(a))}{cos^2(a) - sin^2(a) }[/tex]
Apply difference of two squares to the denominator
[tex]\frac{(cos(a) + sin(a))(cos(a)+ sin(a))}{(cos(a) - sin(a))(cos(a) + sin(a)) }[/tex]
Divide the numerator and denominator by [tex]cos(a) + sin(a)[/tex]
[tex]\frac{cos(a)+ sin(a)}{cos(a) - sin(a) }[/tex]
Hence:
[tex]\frac{1 + sin(2 a)}{cos(2 a) } = \frac{cos(a)+ sin(a)}{cos(a) - sin(a) }[/tex]
