Inverse Trigonometric Functions Question 92
Question: If $ {{\sin }^{-1}}x+{{\sin }^{-1}}y=\frac{\pi }{2}, $ then $ \frac{1+x^{4}+y^{4}}{x^{2}-x^{2}y^{2}+y^{2}} $ is equal to
Options:
A) 1
B) 2
C) ½
D) None of these
Show Answer
Answer:
Correct Answer: B
$ {{\sin }^{-1}}x+{{\sin }^{-1}}y=\frac{\pi }{2}\Rightarrow {{\sin }^{-1}}x=\frac{\pi }{2}-{{\sin }^{-1}}y $
$ \Rightarrow {{\sin }^{-1}}x={{\sin }^{-1}}\sqrt{1-y^{2}}\Rightarrow x^{2}+y^{2}=1 $
$ \Rightarrow \frac{1+x^{4}+y^{4}}{x^{2}-x^{2}y^{2}+y^{2}}=\frac{1+{{(x^{2}+y^{2})}^{2}}-2x^{2}y^{2}}{1-x^{2}y^{2}} $
$ =\frac{1+1-2x^{2}y^{2}}{1-x^{2}y^{2}}=2. $
BETA
