A) \[\frac{1}{\sqrt{3}}\]
B) \[\frac{1}{3}\]
C) \[\frac{1}{6}\]
D) \[\frac{1}{\sqrt{6}}\]
Correct Answer: C
Solution :
Given, \[{{\tan }^{-1}}\,2\theta +{{\tan }^{-1}}\,3\theta =\frac{\pi }{4}\] \[\Rightarrow \] \[{{\tan }^{-1}}\left( \frac{2\theta +3\theta }{1-2\theta \times 3\theta } \right)=\frac{\pi }{4}\] \[\Rightarrow \] \[\frac{5\theta }{1-6{{\theta }^{2}}}=1\] \[\Rightarrow \] \[6{{\theta }^{2}}+5\theta -1=0\] \[\Rightarrow \] \[\theta =\frac{-5\pm \sqrt{25+24}}{2\times 6}\] \[=\frac{-5\pm 7}{12}=-1,\frac{1}{6}\] \[\Rightarrow \] \[\theta =\frac{1}{6}\]
You need to login to perform this action.
You will be redirected in
3 sec
