A) \[2+\sqrt{3}\]
B) \[2-\sqrt{3}\]
C) \[1\]
D) \[\sqrt{3}\]
Correct Answer: B
Solution :
[b] \[\tan (2A-3B).\tan (4B-A)=1,\,\,A,\,\,B\in \left( 0,\frac{\pi }{2} \right)\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\tan (2A-3B)=\cot \,(4B-A)\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\frac{\sin (2A-3B)}{\cos (2A-3B)}=\frac{\cos (4B-A)}{\sin (4B-A)}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\cos \,(4B-A)\,\cos (2A-3B)\] \[-\sin \,(2A-3B)\sin (4B-A)=0\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cos (B+A)=0\] \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,B+A=\frac{\pi }{2}\] \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\tan \left( \frac{B+A}{6} \right)=\tan \frac{\pi }{12}=2-\sqrt{3}\]
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