Consider equation $\sin ^{-1}(px)+ \cos^{-1} (qxy)= sin ^{-1} y$. Let $L$ is the length of locus of point $A(x,y)$ when $p=1=q+1$. Then what is the value of $A$.
My approach :
We have $p=1$ and $q=0$ Hence the equation turns to $$\sin ^{-1}(x)+ \frac {\pi}{2}= sin ^{-1} y$$ Taking sine both sides The equation turns to $$\cos(\sin ^{-1} x)=y$$ Hence $$\cos(\cos ^{-1} \sqrt {1-x^2})=y$$ Hence $$\sqrt {1-x^2}=y$$ So we get locus as $$x^2+y^2=1$$ Hence The length of this locus (circle in this case) is $2\pi$
But the answer is given as $\frac {\pi}{2}$
As $x$ takes all values in $[-1,1]$, the left-hand side of your equation varies from $0$ to $\pi$. But the left-hand side cannot exceed $\pi/2$, hence $x$ is bounded to $[-1,0]$ and $y=\sqrt{1-x^2}\in[0,1]$. The locus is then a quarter circle and its length is $\pi/2$.