Question

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

Solution

## The correct option is C Assertion is correct but Reason is incorrectReason 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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