Inverse Trigonometric Functions Question 150
Question: $ {{\tan }^{-1}}( \frac{x}{y} )-{{\tan }^{-1}}( \frac{x-y}{x+y} ) $ is
[EAMCET 1992]
Options:
A) $ \frac{\pi }{2} $
B) $ \frac{\pi }{3} $
C) $ \frac{\pi }{4} $
D) $ \frac{\pi }{4} $ or $ -\frac{3\pi }{4} $
Show Answer
Answer:
Correct Answer: C
Solution:
$ {{\tan }^{-1}}\frac{x}{y}-{{\tan }^{-1}}( \frac{x-y}{x+y} )={{\tan }^{-1}}\frac{x}{y}-{{\tan }^{-1}}( \frac{1-y/x}{1+y/x} ) $
$ ={{\tan }^{-1}}\frac{x}{y}-( {{\tan }^{-1}}1-{{\tan }^{-1}}\frac{y}{x} ) $
$ ={{\tan }^{-1}}\frac{x}{y}+{{\tan }^{-1}}\frac{y}{x}-\frac{\pi }{4} $
$ ={{\tan }^{-1}}\frac{x}{y}+{{\cot }^{-1}}\frac{x}{y}-\frac{\pi }{4}=\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4} $ .
BETA
