Inverse Trigonometric Functions Question 40
Question: If $ \angle A=90^{o} $ in the triangle ABC, then $ {{\tan }^{-1}}( \frac{c}{a+b} )+{{\tan }^{-1}}( \frac{b}{a+c} )= $
[Kerala (Engg.) 2005]
Options:
A) 0
B) 1
C) $ \pi /4 $
D) $ \pi /6 $
Show Answer
Answer:
Correct Answer: C
Solution:
$ \angle A=90^{o} $
$ {{\tan }^{-1}}( \frac{c}{a+b} )+{{\tan }^{-1}}( \frac{b}{a+c} ) $
$ ={{\tan }^{-1}}[ \frac{\frac{c}{a+b}+\frac{b}{a+c}}{1-( \frac{c}{a+b} )( \frac{b}{a+c} )} ] $
$ ={{\tan }^{-1}}[ \frac{ca+c^{2}+ab+b^{2}}{a^{2}+ab+ca+bc-bc} ] $
$ ={{\tan }^{-1}}[ \frac{a^{2}+ab+ca}{a^{2}+ab+ca} ] $
$ ={{\tan }^{-1}}(1)=\frac{\pi }{4} $ .
BETA
