Inverse Trigonometric Functions Question 211
Question: If $ x^{2}+y^{2}+z^{2}=r^{2} $ , then $ {{\tan }^{-1}}( \frac{xy}{zr} )+ $
$ {{\tan }^{-1}}( \frac{yz}{xr} )+\tan ( \frac{zx}{yr} )= $
Options:
A) $ \pi $
B) $ \frac{\pi }{2} $
C) 0
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ {{\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
