Question

If $si{n}^{-1}x+si{n}^{-1}y=\frac{\pi }{2},then\frac{1+x^4+y^4}{x^2-x^2{y}^2+y^2}$ is equal to

• A. $1$
• B. $2$
• C. $\frac{1}{2}$
• D. $noneofthese$

Answer 1

The correct option is B $2$

$si{n}^{-1}x+si{n}^{-1}y=\frac{\pi }{2}\Rightarrow si{n}^{-1}(x\sqrt{1-y^2}+y\sqrt{1-x^2})=\frac{\pi }{2}\Rightarrow x\sqrt{1-y^2}+y\sqrt{1-x^2}=1\Rightarrow x\sqrt{1-y^2}=1-y\sqrt{1-x^2}\Rightarrow x^2(1-y^2)=1+y^2(1-x^2)-2y\sqrt{1-x^2}\Rightarrow x^2-x^2{y}^2=1+y^2-x^2{y}^2-2y\sqrt{1-x^2}\Rightarrow x^2-y^2-1=-2y\sqrt{1-x^2}\Rightarrow x^4+y^4+1-2x^2{y}^2-2x^2+2y^2=4y^2(1-x^2)\Rightarrow x^4+y^4+1=2(x^2-x^2{y}^2+y^2)\Rightarrow \frac{x^4+y^4+1}{x^2-x^2{y}^2+y^2}=2$