Inverse Trigonometric Functions Question 114
Question: If $ {{\sin }^{-1}}x+{{\sin }^{-1}}y+{{\sin }^{-1}}z=\pi , $ then $ x^{4}+y^{4}+z^{4}+4x^{2}y^{2}z^{2}=k(x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2}). $ where k =
Options:
A) 1
B) 2
C) 4
D) None of these
Show Answer
Answer:
Correct Answer: B
We have, $ {{\sin }^{-1}}x+{{\sin }^{-1}}y=\pi -{{\sin }^{-1}}z $
or $ x\sqrt{(1-y^{2})}+y\sqrt{(1-x^{2})}=z $
or $ x^{2}(1-y^{2})=z^{2}+y^{2}(1-x^{2})-2yz\sqrt{(1-x^{2})} $
or $ {{(x^{2}-z^{2}-y^{2})}^{2}}=4y^{2}z^{2}(1-x^{2}) $
or $ x^{4}+y^{4}+z^{4}-2x^{2}z^{2}+2y^{2}z^{2}-2x^{2}y^{2} $
$ +4x^{2}y^{2}z^{2}-4y^{2}z^{2}=0 $
or $ x^{4}+y^{4}+z^{4}+4x^{2}y^{2}z^{2}=2(x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2}) $
$ \therefore k=2 $
BETA
