A) \[a+b=c\]
B) \[b+c=a\]
C) \[a+c=b\]
D) \[b=c\]
Correct Answer: A
Solution :
Given \[\tan \frac{P}{2}+\tan \frac{Q}{2}=-\frac{b}{a}\] and \[\tan \frac{P}{2}\tan \frac{Q}{2}=\frac{c}{a}\] or \[\tan \frac{\alpha }{2}+\tan \left( \frac{\pi }{4}-\frac{\alpha }{2} \right)=-\frac{b}{a}\],\[\tan \frac{\alpha }{2}.\tan \left( \frac{\pi }{4}-\frac{\alpha }{2} \right)=\frac{c}{a}\] \[\therefore \]\[\tan \frac{\alpha }{2}+\frac{1-\tan \frac{\alpha }{2}}{1+\tan \frac{\alpha }{2}}=-\frac{b}{a}\]\[\Rightarrow \]\[\frac{{{\tan }^{2}}\frac{\alpha }{2}+1}{1+\tan \frac{\alpha }{2}}=-\frac{b}{a}\] ?..(i) Similarly, \[\frac{\tan \frac{\alpha }{2}\left( 1-\tan \frac{\alpha }{2} \right)}{1+\tan \frac{\alpha }{2}}=\frac{c}{a}\] \[\Rightarrow \] \[\frac{\tan \frac{\alpha }{2}-{{\tan }^{2}}\frac{\alpha }{2}}{1+\tan \frac{\alpha }{2}}=\frac{c}{a}\] ......(ii) By adding (i) and (ii), we get \[\frac{1+\tan \frac{\alpha }{2}}{1+\tan \frac{\alpha }{2}}=-\frac{b}{a}+\frac{c}{a}\] \[\Rightarrow \] \[-b+c=a\] \[\Rightarrow \] \[c=a+b\].
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