A) \[\frac{c+a}{2b}\]
B) \[\frac{2b}{c+a}\]
C) \[\frac{c-a}{2b}\]
D) \[\frac{b}{c+a}\]
Correct Answer: B
Solution :
\[a\cos 2\theta +b\sin 2\theta =c\] Þ \[a\left( \frac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta } \right)+b\frac{2\tan \theta }{1+{{\tan }^{2}}\theta }=c\] \[\Rightarrow \] \[a-a{{\tan }^{2}}\theta +2b\tan \theta =c+c{{\tan }^{2}}\theta \] \[\Rightarrow \]\[-(a+c){{\tan }^{2}}\theta +2b\,\tan \theta +(a-c)=0\] \[\therefore \tan \alpha +\tan \beta =-\frac{2b}{-(c+a)}=\frac{2b}{c+a}\] .
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