A) 1/3
B) 1/4
C) 3
D) 4
Correct Answer: C
Solution :
We have \[{{\cot }^{-1}}x+{{\tan }^{-1}}3=\frac{\pi }{2}\] \[\Rightarrow \,\,{{\cot }^{-1}}x+{{\tan }^{-1}}3=\frac{\pi }{2}\,\,\Rightarrow \,\,{{\tan }^{-1}}\frac{1}{x}+{{\tan }^{-1}}3=\frac{\pi }{2}\] \[\Rightarrow \,\,{{\tan }^{-1}}\left( \frac{\frac{1}{x}+3}{1-\frac{1}{x}.3} \right)={{\tan }^{-1}}\left( \frac{1}{0} \right)\] \[\Rightarrow \,\,\frac{3x+1}{x-3}=\frac{1}{0}\,\,\Rightarrow \,\,x=3\] Aliter: As we know that, \[{{\tan }^{-1}}x+{{\cot }^{-1}}x=\frac{\pi }{2},\] therefore for the given question, x should be 3.
You need to login to perform this action.
You will be redirected in
3 sec
