Inverse Trigonometric Functions Question 227
Question: $ {{\tan }^{-1}}( \frac{1}{11} )+{{\tan }^{-1}}( \frac{2}{12} )= $
[DCE 1999]
Options:
A) $ {{\tan }^{-1}}( \frac{33}{132} ) $
B) $ {{\tan }^{-1}}( \frac{1}{2} ) $
C) $ {{\tan }^{-1}}( \frac{132}{33} ) $
D) None of these
Show Answer
Answer:
Correct Answer: D
Solution:
$ {{\tan }^{-1}}( \frac{1}{11} )+{{\tan }^{-1}}( \frac{2}{12} ) $ = $ {{\tan }^{-1}}( \frac{\frac{1}{11}+\frac{2}{12}}{1-\frac{1}{11}\times \frac{2}{12}} )={{\tan }^{-1}}( \frac{12+22}{130} ) $ = $ {{\tan }^{-1}}( \frac{34}{130} )={{\tan }^{-1}}( \frac{17}{65} ) $ .
BETA
