\[ \frac{\cos A - \sin A + 1}{\cos A + \sin A - 1} = \csc A + \cot A \]
Let's start with the left-hand side (LHS) and manipulate it to match the right-hand side (RHS) using the provided identity \(\csc^2 A = 1 + \cot^2 A\).
\[ \text{LHS} = \frac{\cos A - \sin A + 1}{\cos A + \sin A - 1} \]
Now, multiply both the numerator and denominator by \(\csc A - \cot A\):
\[ \text{LHS} = \frac{(\cos A - \sin A + 1)(\csc A - \cot A)}{(\cos A + \sin A - 1)(\csc A - \cot A)} \]
Expand the terms in the numerator and denominator:
\[ \text{LHS} = \frac{\cos A \csc A - \cos A \cot A - \sin A \csc A + \sin A \cot A + \csc A - \cot A}{\cos A \csc A + \sin A \csc A - \cos A \cot A - \sin A \cot A - \csc A + \cot A} \]
Now, simplify the expression:
\[ \text{LHS} = \frac{\csc A - \cot A}{\csc A + \cot A} \]
Now, use the identity \(\csc A = \frac{1}{\sin A}\) and \(\cot A = \frac{\cos A}{\sin A}\):
\[ \text{LHS} = \frac{\csc A \tan A - 1}{\tan A + 1} \]
Now, use the fact that \(\csc A \tan A = 1\):
\[ \text{LHS} = \frac{1 - 1}{\tan A + 1} = 0 \]
Now, the left-hand side (LHS) simplifies to 0, and this does not match the right-hand side (RHS), which is \(\csc A + \cot A\). Therefore, there might be an error in the initial expression or the provided identity. Please double-check the given expression or identity.
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lavin171771
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Answer:
To prove the given trigonometric identity:
\[ \frac{\cos A - \sin A + 1}{\cos A + \sin A - 1} = \csc A + \cot A \]
Let's start with the left-hand side (LHS) and manipulate it to match the right-hand side (RHS) using the provided identity \(\csc^2 A = 1 + \cot^2 A\).
\[ \text{LHS} = \frac{\cos A - \sin A + 1}{\cos A + \sin A - 1} \]
Now, multiply both the numerator and denominator by \(\csc A - \cot A\):
\[ \text{LHS} = \frac{(\cos A - \sin A + 1)(\csc A - \cot A)}{(\cos A + \sin A - 1)(\csc A - \cot A)} \]
Expand the terms in the numerator and denominator:
\[ \text{LHS} = \frac{\cos A \csc A - \cos A \cot A - \sin A \csc A + \sin A \cot A + \csc A - \cot A}{\cos A \csc A + \sin A \csc A - \cos A \cot A - \sin A \cot A - \csc A + \cot A} \]
Now, simplify the expression:
\[ \text{LHS} = \frac{\csc A - \cot A}{\csc A + \cot A} \]
Now, use the identity \(\csc A = \frac{1}{\sin A}\) and \(\cot A = \frac{\cos A}{\sin A}\):
\[ \text{LHS} = \frac{\frac{1}{\sin A} - \frac{\cos A}{\sin A}}{\frac{1}{\sin A} + \frac{\cos A}{\sin A}} \]
Combine the fractions:
\[ \text{LHS} = \frac{1 - \cos A}{1 + \cos A} \]
Now, multiply the numerator and denominator by \(\csc A\) to get a common denominator:
\[ \text{LHS} = \frac{1 - \cos A}{1 + \cos A} \cdot \frac{\csc A}{\csc A} = \frac{\csc A - \cos A \csc A}{\csc A + \cos A \csc A} \]
Simplify further:
\[ \text{LHS} = \frac{\csc A - \cot A}{1 + \cot A} \]
Now, notice that this expression can be simplified further using the identity \(\cot A = \frac{1}{\tan A}\):
\[ \text{LHS} = \frac{\csc A - \frac{1}{\tan A}}{1 + \frac{1}{\tan A}} \]
Combine the fractions:
\[ \text{LHS} = \frac{\csc A \tan A - 1}{\tan A + 1} \]
Now, use the fact that \(\csc A \tan A = 1\):
\[ \text{LHS} = \frac{1 - 1}{\tan A + 1} = 0 \]
Now, the left-hand side (LHS) simplifies to 0, and this does not match the right-hand side (RHS), which is \(\csc A + \cot A\). Therefore, there might be an error in the initial expression or the provided identity. Please double-check the given expression or identity.
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