A) \[\pi /3\]
B) \[2\pi /3\]
C) \[\pi /2\]
D) \[3\pi /4\]
Correct Answer: C
Solution :
\[2\tan x-k(1+ta{{n}^{2}}x)=0\,\,;\,\,0<k<1\] |
Since \[\angle B\]and \[\angle C\]of a \[\Delta ABC\] satisfy it |
\[\therefore \tan C+\tan B\,\,are\,\,roots\]\[\therefore \,\,\,\tan C+\tan B=\frac{2}{K}\] and \[\tan C\tan B=1\] |
\[\Rightarrow \]\[\tan (C+B)=\frac{\tan C+\tan B}{1-\tan C\tan B}=\frac{\frac{2}{k}}{1-1}\]\[\Rightarrow B+C=\frac{\pi }{2}\,\,\,\,\therefore \,\,\,\,\,A=\frac{\pi }{2}\] |
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