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Assertion :$$\dfrac {\sin (A + B) + \sin (A - B)}{\cos (A + B) + \cos (A - B)} = \tan {A}$$ Reason: $$\sin (A + B) + \sin (A - B) = \sin {A}$$ and $$\cos (A + B) + \cos (A - B) = \cos {A}$$


A
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
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B
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
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C
Assertion is correct but Reason is incorrect
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D
Assertion is incorrect but Reason is correct
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Solution

The correct option is C Assertion is correct but Reason is incorrect
Reason is false as

$$\sin (A + B) + \sin (A - B)$$

$$= 2 \sin \left (\dfrac {(A + B) + (A - B)}{2}\right ) \cos \left (\dfrac {(A + B) - (A - B)}{2}\right )$$

$$= 2 \sin {A} \cos {B}$$

And

$$\cos (A + B) + \cos (A - B)$$

$$= 2 \cos \left (\dfrac {(A + B) + (A - B)}{2}\right ) \cos \left (\dfrac {(A + B) - (A - B)}{2}\right )$$

$$= 2 \cos {A} \cos {B}$$

Assertion is true as

$$\dfrac {\sin (A + B) + \sin  (A - B)}{\cos (A + B) + \cos (A - B)} = \dfrac {2\sin {A} \cos {B}}{2 \cos {A} \cos {B}} =\dfrac{\sin A}{\cos A}= \tan {A}$$

Mathematics

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