Certainly! To prove the trigonometric identity [tex]\(\frac{1 - \tan a}{1 + \tan a} = \frac{1 - \sin 2a}{\cos 2a}\)[/tex], let's start by transforming both sides and showing their equivalence step-by-step.
### Left Hand Side (LHS):
[tex]\[ \frac{1 - \tan a}{1 + \tan a} \][/tex]
We know the identity for tangent:
[tex]\[ \tan a = \frac{\sin a}{\cos a} \][/tex]
Substitute [tex]\(\tan a = \frac{\sin a}{\cos a}\)[/tex] into the LHS:
[tex]\[ \frac{1 - \frac{\sin a}{\cos a}}{1 + \frac{\sin a}{\cos a}} \][/tex]
To simplify, multiply the numerator and the denominator by [tex]\(\cos a\)[/tex]:
[tex]\[ \frac{\cos a \left(1 - \frac{\sin a}{\cos a}\right)}{\cos a \left(1 + \frac{\sin a}{\cos a}\right)} = \frac{\cos a - \sin a}{\cos a + \sin a} \][/tex]
So the LHS becomes:
[tex]\[ \frac{\cos a - \sin a}{\cos a + \sin a} \][/tex]
### Right Hand Side (RHS):
[tex]\[ \frac{1 - \sin 2a}{\cos 2a} \][/tex]
We use the double-angle identities for sine and cosine:
[tex]\[ \sin 2a = 2 \sin a \cos a \][/tex]
[tex]\[ \cos 2a = \cos^2 a - \sin^2 a \][/tex]
Substitute these identities into the RHS:
[tex]\[ \frac{1 - 2 \sin a \cos a}{\cos^2 a - \sin^2 a} \][/tex]
We recognize that we can reframe both numerator and denominator in terms of [tex]\((\cos a - \sin a)\)[/tex]:
Consider squaring [tex]\((\cos a - \sin a)\)[/tex] and [tex]\((\cos a + \sin a)\)[/tex]:
[tex]\[ (\cos a - \sin a)^2 = \cos^2 a - 2 \sin a \cos a + \sin^2 a \][/tex]
[tex]\[ (\cos a + \sin a)^2 = \cos^2 a + 2 \sin a \cos a + \sin^2 a \][/tex]
Notice these can help us with both numerator and denominator:
The numerator:
[tex]\[ 1 - 2 \sin a \cos a = (\cos^2 a + \sin^2 a) - 2 \sin a \cos a = (\cos a - \sin a)^2 \][/tex]
The denominator is a standard double-angle identity:
[tex]\[ \cos^2 a - \sin^2 a = (\cos a + \sin a)(\cos a - \sin a) \][/tex]
Therefore,
[tex]\[ \frac{1 - 2 \sin a \cos a}{\cos^2 a - \sin^2 a} = \frac{(\cos a - \sin a)^2}{(\cos a + \sin a)(\cos a - \sin a)} \][/tex]
Simplify the right side:
[tex]\[ \frac{(\cos a - \sin a)^2}{(\cos a + \sin a)(\cos a - \sin a)} = \frac{(\cos a - \sin a)}{(\cos a + \sin a)} \][/tex]
### Conclusion:
Both sides simplify to the same expression:
[tex]\[ \frac{\cos a - \sin a}{\cos a + \sin a} \][/tex]
Thus, the identity is proven to be valid:
[tex]\[ \frac{1 - \tan a}{1 + \tan a} = \frac{1 - \sin 2a}{\cos 2a} \][/tex]
