| If \[(1+\tan A)(1+\tan B)=2,\]then \[(A+B)\]is equal to |
A) \[\frac{\pi }{2}\]
B) \[\frac{\pi }{3}\]
C) \[\frac{\pi }{4}\]
D) \[\frac{\pi }{6}\]
E) None of these
Correct Answer: C
Solution :
| Given, \[1+\tan A+\tan B+\tan A\tan B=2\] |
| \[\Rightarrow \]\[\tan A+\tan B=1-\tan A\tan B\] |
| \[\Rightarrow \]\[\frac{\tan A+\tan B}{1-\tan \,\,A\,\,tan\,\,B}=1=\tan 45{}^\circ \] |
| \[\Rightarrow \]\[\tan \,\,(A+B)=\tan 45{}^\circ \] |
| \[\left[ \because \tan \,\,(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B} \right]\] |
| \[\therefore \]\[A+B=45{}^\circ =\frac{\pi }{4}\] |
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