Inverse Trigonometric Functions Question 74

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) $ \frac{1}{2} $

D) none of these

Show Answer

Answer:

Correct Answer: B

$ {{\sin }^{-1}}x+{{\sin }^{-1}}y=\frac{\pi }{2} $

Or $ {{\sin }^{-1}}x=\frac{\pi }{2}-{{\sin }^{-1}}y $

Or $ {{\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 $