A) \[\frac{a}{b}\]
B) \[\frac{b}{a}\]
C) \[\frac{c}{a}\]
D) \[\frac{a}{c}\]
Correct Answer: B
Solution :
\[\cos x+b\sin x=c\] \[\Rightarrow \] \[a\frac{\left( 1-{{\tan }^{2}}\frac{x}{2} \right)}{\left( 1+{{\tan }^{2}}\frac{x}{2} \right)}+\frac{2b\tan \frac{x}{2}}{1+{{\tan }^{2}}\frac{x}{2}}=c\] \[\Rightarrow \] \[(c+a){{\tan }^{2}}\frac{x}{2}-2b\tan \frac{x}{2}+c-a=0\] \[\Rightarrow \] \[\tan \frac{{{x}_{1}}}{2}+\tan \frac{{{x}_{2}}}{2}=\frac{2b}{c+a}\] \[\tan \frac{{{x}_{1}}}{2}\cdot \tan \frac{{{x}_{2}}}{2}=\frac{c-a}{c+a}\] \[\Rightarrow \] \[\tan \left( \frac{{{x}_{1}}+{{x}_{2}}}{2} \right)=\frac{\frac{2b}{c+a}}{1-\frac{c-a}{c+a}}\] \[=\frac{2b}{2a}=\frac{b}{a}\]
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