Trigonometric Identities Question 349
Question: If $ \sin A+\sin B=C,\cos A+\cos B=D, $ then the value of $ \sin (A+B)= $
[MP PET 1986]
Options:
A) $ CD $
B) $ \frac{CD}{C^{2}+D^{2}} $
C) $ \frac{C^{2}+D^{2}}{2,CD} $
D) $ \frac{2,CD}{C^{2}+D^{2}} $
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Answer:
Correct Answer: D
Solution:
As given $ \frac{\sin A+\sin B}{\cos A+\cos B}=\frac{C}{D} $
$ \Rightarrow \frac{2\sin \frac{A+B}{2}.\cos \frac{A-B}{2}}{2\cos \frac{A+B}{2}.\cos \frac{A-B}{2}}=\frac{C}{D} $
$ \Rightarrow \tan \frac{A+B}{2}=\frac{C}{D} $ Thus, $ \sin ,(A+B)=\frac{2\tan \frac{A+B}{2}}{1+{{\tan }^{2}}\frac{A+B}{2}} $ $ =\frac{2,\frac{C}{D}}{1+\frac{C^{2}}{D^{2}}}=\frac{2CD}{(C^{2}+D^{2})} $ .
BETA
