Inverse Trigonometric Functions Question 175
Question: If $ {{\sin }^{-1}}x+{{\sin }^{-1}}y+{{\sin }^{-1}}z=\frac{\pi }{2} $ , then the value of $ x^{2}+y^{2}+z^{2}+2xyz $ is equal to
[Pb. CET 2002]
Options:
A) 0
B) 1
C) 2
D) 3
Show Answer
Answer:
Correct Answer: B
Solution:
$ {{\sin }^{-1}}x+{{\sin }^{-1}}y+{{\sin }^{-1}}z=\frac{\pi }{2} $ Put $ {{\sin }^{-1}}x=\alpha $ , $ {{\sin }^{-1}}y=\beta , $
$ {{\sin }^{-1}}z=\gamma $
$ \therefore $ $ \alpha +\beta +\gamma =\frac{\pi }{2} $ , (Given) or $ \alpha +\beta =\frac{\pi }{2}-\gamma $
Or $ \cos (\alpha +\beta )=\cos ( \frac{\pi }{2}-\gamma ) $
$ \cos \alpha \cos \beta -\sin \alpha \sin \beta =\sin \gamma $ -. .(i) and, we have $ \sin \alpha =x $
Therefore $ \cos \alpha =\sqrt{1-x^{2}} $ Similarly, $ \cos \beta =\sqrt{1-y^{2}} $
$ \therefore $ From equation (i), we get $ \sqrt{1-x^{2}}.\sqrt{1-y^{2}}=xy+z $ Squaring both sides, we have $ x^{2}+y^{2}+z^{2}+2xyz=1 $ .
BETA
