Question

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

Answer 1

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

Reason is false as

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

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

$=2\sin A\cos B$

And

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

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

$=2\cos A\cos B$

Assertion is true as

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