Answer :
To solve the given trigonometric identity \(\frac{1 - \cos 2A + \sin 2A}{1 + \cos 2A + \sin 2A} = \tan A\), let's follow a series of steps involving trigonometric identities and simplifications.
1. Recall Double Angle Identities:
- \(\cos 2A = \cos^2 A - \sin^2 A\)
- \(\sin 2A = 2 \sin A \cos A\)
2. Substitute these identities into the expression:
Given the expression:
[tex]\[ \frac{1 - \cos 2A + \sin 2A}{1 + \cos 2A + \sin 2A} \][/tex]
Replace \(\cos 2A\) and \(\sin 2A\) with the double angle identities:
[tex]\[ \frac{1 - (\cos^2 A - \sin^2 A) + 2 \sin A \cos A}{1 + (\cos^2 A - \sin^2 A) + 2 \sin A \cos A} \][/tex]
3. Simplify the numerator:
Distribute and combine like terms in the numerator:
[tex]\[ 1 - \cos^2 A + \sin^2 A + 2 \sin A \cos A \][/tex]
Recognize that \(1 - \cos^2 A = \sin^2 A\), so:
[tex]\[ \sin^2 A + \sin^2 A + 2 \sin A \cos A = 2 \sin^2 A + 2 \sin A \cos A \][/tex]
4. Simplify the denominator:
Distribute and combine like terms in the denominator:
[tex]\[ 1 + \cos^2 A - \sin^2 A + 2 \sin A \cos A \][/tex]
Recognize that \(1 + \cos^2 A - \sin^2 A = \cos^2 A + \cos^2 A = 2 \cos^2 A\),
so:
[tex]\[ 2 \cos^2 A + 2 \sin A \cos A \][/tex]
5. Rewrite the expression:
[tex]\[ \frac{2 \sin^2 A + 2 \sin A \cos A}{2 \cos^2 A + 2 \sin A \cos A} \][/tex]
6. Factor out common terms:
Factor \(2\) from the numerator and the denominator:
[tex]\[ \frac{2 (\sin^2 A + \sin A \cos A)}{2 (\cos^2 A + \sin A \cos A)} \][/tex]
Cancel the common factor \(2\):
[tex]\[ \frac{\sin^2 A + \sin A \cos A}{\cos^2 A + \sin A \cos A} \][/tex]
7. Simplify further:
Recognize \(\sin A \cos A\) is common in both the numerator and the denominator and factor it out:
[tex]\[ \frac{\sin A (\sin A + \cos A)}{\cos A (\cos A + \sin A)} \][/tex]
The \((\sin A + \cos A)\) terms cancel each other out:
[tex]\[ \frac{\sin A}{\cos A} = \tan A \][/tex]
Hence, we have successfully shown that:
[tex]\[ \frac{1 - \cos 2A + \sin 2A}{1 + \cos 2A + \sin 2A} = \tan A \][/tex]
Therefore, the trigonometric identity is correct.
1. Recall Double Angle Identities:
- \(\cos 2A = \cos^2 A - \sin^2 A\)
- \(\sin 2A = 2 \sin A \cos A\)
2. Substitute these identities into the expression:
Given the expression:
[tex]\[ \frac{1 - \cos 2A + \sin 2A}{1 + \cos 2A + \sin 2A} \][/tex]
Replace \(\cos 2A\) and \(\sin 2A\) with the double angle identities:
[tex]\[ \frac{1 - (\cos^2 A - \sin^2 A) + 2 \sin A \cos A}{1 + (\cos^2 A - \sin^2 A) + 2 \sin A \cos A} \][/tex]
3. Simplify the numerator:
Distribute and combine like terms in the numerator:
[tex]\[ 1 - \cos^2 A + \sin^2 A + 2 \sin A \cos A \][/tex]
Recognize that \(1 - \cos^2 A = \sin^2 A\), so:
[tex]\[ \sin^2 A + \sin^2 A + 2 \sin A \cos A = 2 \sin^2 A + 2 \sin A \cos A \][/tex]
4. Simplify the denominator:
Distribute and combine like terms in the denominator:
[tex]\[ 1 + \cos^2 A - \sin^2 A + 2 \sin A \cos A \][/tex]
Recognize that \(1 + \cos^2 A - \sin^2 A = \cos^2 A + \cos^2 A = 2 \cos^2 A\),
so:
[tex]\[ 2 \cos^2 A + 2 \sin A \cos A \][/tex]
5. Rewrite the expression:
[tex]\[ \frac{2 \sin^2 A + 2 \sin A \cos A}{2 \cos^2 A + 2 \sin A \cos A} \][/tex]
6. Factor out common terms:
Factor \(2\) from the numerator and the denominator:
[tex]\[ \frac{2 (\sin^2 A + \sin A \cos A)}{2 (\cos^2 A + \sin A \cos A)} \][/tex]
Cancel the common factor \(2\):
[tex]\[ \frac{\sin^2 A + \sin A \cos A}{\cos^2 A + \sin A \cos A} \][/tex]
7. Simplify further:
Recognize \(\sin A \cos A\) is common in both the numerator and the denominator and factor it out:
[tex]\[ \frac{\sin A (\sin A + \cos A)}{\cos A (\cos A + \sin A)} \][/tex]
The \((\sin A + \cos A)\) terms cancel each other out:
[tex]\[ \frac{\sin A}{\cos A} = \tan A \][/tex]
Hence, we have successfully shown that:
[tex]\[ \frac{1 - \cos 2A + \sin 2A}{1 + \cos 2A + \sin 2A} = \tan A \][/tex]
Therefore, the trigonometric identity is correct.
