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e. [tex]\[\frac{1 - \tan a}{1 + \tan a} = \frac{1 - \sin 2a}{\cos 2a}\][/tex]

Sagot :

GinnyAnswer
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]
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