Question

If ${\sin }^{-1}x+{\sin }^{-1}y=\frac{\pi }{2}$,then p$(x^2-x^2{y}^2+y^2)=q+x^4+y^4$
Find the value of $p+q$.

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is C $3$

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

$\Rightarrow {\sin }^{-1}[x\sqrt{1-y^2}+y\sqrt{1-x^2}]=\frac{\pi }{2}$

$\Rightarrow [x\sqrt{1-y^2}+y\sqrt{1-x^2}]=\sin \frac{\pi }{2}=1$

$\Rightarrow x\sqrt{1-y^2}=1-y\sqrt{1-x^2}$.

Squaring both sides, $x^2(1-y^2)=1+y^2(1-x^2)-2y\sqrt{1-x^2}$

$\Rightarrow 1+y^2-x^2=2y\sqrt{1-x^2}$

Again squaring, $1+y^4+x^4+2y^2-2x^2-2x^2{y}^2=4y^2(1-x^2)$

$\Rightarrow 2(y^2+x^2-x^2{y}^2)=1+x^4+y^4$

by comparing it with $p(x^2-x^2{y}^2+y^2)=q+x^4+y^4$

we get $p=2,q=1$

$\therefore p+q=2+1=3$