Inverse Trigonometric Functions Question 203
Question: If $ {{\cot }^{-1}}x+{{\tan }^{-1}}3=\frac{\pi }{2} $ , then x =
Options:
A) 1/3
B) 1/4
C) 3
D) 4
Show Answer
Answer:
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}}( \frac{\frac{1}{x}+3}{1-\frac{1}{x}.3} )={{\tan }^{-1}}( \frac{1}{0} ) $
$ \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.
BETA
