A) \[a+b\]
B) \[a-b\]
C) \[\frac{b}{a}\]
D) \[\frac{a}{b}\]
Correct Answer: C
Solution :
\[a\cos x+b\sin x=c\] \[\Rightarrow \] \[a\,\left( \frac{1-{{\tan }^{2}}\,(x/2)}{1+{{\tan }^{2}}(x/2)} \right)+\frac{2b\tan (x/2)}{1+{{\tan }^{2}}(x/2)}=c\] \[\Rightarrow \] \[(a+c){{\tan }^{2}}\frac{x}{2}-2b\tan \frac{x}{2}+(c-a)=0\] This equation has roots \[\tan \frac{\alpha }{2}\] and \[\tan \frac{\beta }{2}\]. \[\therefore \] \[\tan \frac{\alpha }{2}+\tan \frac{\beta }{2}=\frac{2b}{a+c}\] and \[\tan \frac{\alpha }{2}\tan \frac{\beta }{2}=\frac{c-a}{a+c}\] Now\[\tan \left( \frac{\alpha }{2}+\frac{\beta }{2} \right)=\frac{\tan \frac{\alpha }{2}+\tan \frac{\beta }{2}}{1-\tan \frac{\alpha }{2}\tan \frac{\beta }{2}}=\frac{\frac{2b}{a+c}}{1-\frac{c-a}{a+c}}=\frac{b}{a}\].
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