Inverse Trigonometric Functions Question 110
Question: If $ x^{2}+y^{2}+z^{2}=r^{2}, $ then $ {{\tan }^{-1}}\frac{xy}{zr}+{{\tan }^{-1}}\frac{yz}{xr}+{{\tan }^{-1}}\frac{xz}{yr}= $
Options:
A) $ \pi $
B) $ \frac{\pi }{2} $
C) 0
D) None of these
Show Answer
Answer:
Correct Answer: B
$ {{\tan }^{-1}}\frac{xy}{zr}+{{\tan }^{-1}}\frac{yz}{xr}+{{\tan }^{-1}}\frac{xz}{yr} $
$ ={{\tan }^{-1}}[ \frac{\frac{xy}{zr}+\frac{yz}{xr}+\frac{xz}{yr}-\frac{xyz}{r^{3}}}{1-( \frac{x^{2}+y^{2}+z^{2}}{r^{2}} )} ]={{\tan }^{-1}}\infty =\frac{\pi }{2} $
BETA
