If \( A + B = \frac{\pi}{4} \), then \( \dfrac{\cos B - \sin B}{\cos B + \sin B} \) is equal to:
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- \( \sin A \)
- \( \cos A \)
- \( \tan A \)
- \( \cot A \)
The Correct Option is C
Solution and Explanation
Step 1: We are given \( A + B = \dfrac{\pi}{4} \). This implies \( B = \dfrac{\pi}{4} - A \).
Step 2: Substitute \( B \) into the expression: \[ \dfrac{\cos B - \sin B}{\cos B + \sin B} = \dfrac{\cos(\frac{\pi}{4} - A) - \sin(\frac{\pi}{4} - A)}{\cos(\frac{\pi}{4} - A) + \sin(\frac{\pi}{4} - A)} \] Step 3: Use trigonometric identities: \[ \cos(\frac{\pi}{4} - A) = \cos\frac{\pi}{4}\cos A + \sin\frac{\pi}{4}\sin A = \frac{1}{\sqrt{2}}(\cos A + \sin A) \] \[ \sin(\frac{\pi}{4} - A) = \sin\frac{\pi}{4}\cos A - \cos\frac{\pi}{4}\sin A = \frac{1}{\sqrt{2}}(\cos A - \sin A) \] Step 4: Substitute into the main expression: \[ \dfrac{\frac{1}{\sqrt{2}}(\cos A + \sin A) - \frac{1}{\sqrt{2}}(\cos A - \sin A)}{\frac{1}{\sqrt{2}}(\cos A + \sin A) + \frac{1}{\sqrt{2}}(\cos A - \sin A)} = \dfrac{2\sin A}{2\cos A} = \tan A \]
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