If $ \arcsin x+ \arcsin y=\frac{\pi}{2}$, prove that $ x^2+y^2=1$. If $ \arcsin x+ \arcsin y=\frac{\pi}{2}$,
prove that $$ x^2+y^2=1$$
I tried taking sine of both the sides, I only come to this result: $$x^2 + y^2 -2x^2y^2 + 2xy\sqrt{(1-y^2)(1+x^2)}=1.$$
 A: Take the sine of both sides for a direct demonstration.
$$\sin (\arcsin x+\arcsin y)=1$$
Then $$\sin (\arcsin x)\cos(\arcsin y)+\cos(\arcsin x)+\sin (\arcsin y)=1$$
Now note that $\arcsin y=\frac \pi 2-\arcsin x$ from the original equation and also $\cos \left(\frac \pi2-\theta\right)=\sin \theta$ to obtain $$x \sin(\arcsin x)+y \sin (\arcsin y)=x^2+y^2=1$$
A: Another way to prove it, which might give a little more insight on why this is true, is to look at this triangle:

from Wikimedia Commons
By definition, $\theta$ is equal to the arcsin of $x$; the other sharp angle is $\pi/2 - \theta$ (because the angles of a triangle sum up to $\pi$) and its arcsin is $y = \sqrt{1-x^2}$ (by Pythagoras). Now, arcsin $x$ + arcsin $y$ = $\theta + \pi/2 - \theta$ = $\pi/2$, and you can verify that $x^2 + y^2 = 1$.
A: Let, $\sin^{-1}x=\alpha\iff x=\sin \alpha$ 
& $\sin^{-1}x=\beta\iff y=\sin\beta$ then we have$$\alpha+\beta=\frac{\pi}{2}$$
Now, we have 
$$x^2+y^2=(\sin\alpha)^2+(\sin\beta)^2$$
$$x^2+y^2=\sin^2\alpha+\sin^2\beta$$
Substituting $\beta=\frac{\pi}{2}-\alpha$, we get 
$$x^2+y^2=\sin^2\alpha+\sin^2\left(\frac{\pi}{2}-\alpha\right)$$
$$=\sin^2\alpha+\cos^2\alpha=1$$ 
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{x^2+y^2=1}}$$
A: $\arcsin x=\dfrac\pi2-\arcsin y=\arccos y$
$\arcsin x=\arccos y=A$ (say)
$\implies x=\sin A,y=\cos A$
Hope you can take it from here!
A: $$\sin^{-1}(x)+\sin^{-1}(y)=\frac{\pi}{2}\Longleftrightarrow$$
$$\sin^{-1}(y)=\frac{\pi}{2}-\sin^{-1}(x)\Longleftrightarrow$$
$$\sin(\sin^{-1}(y))=\sin\left(\frac{\pi}{2}-\sin^{-1}(x)\right)\Longleftrightarrow$$
$$y=\sqrt{1-x^2}$$

$$\sin^{-1}(x)+\sin^{-1}(y)=\frac{\pi}{2}\Longleftrightarrow$$
$$\sin^{-1}(x)=\frac{\pi}{2}-\sin^{-1}(y)\Longleftrightarrow$$
$$\sin(\sin^{-1}(x))=\sin\left(\frac{\pi}{2}-\sin^{-1}(y)\right)\Longleftrightarrow$$
$$x=\sqrt{1-y^2}$$

$$x^2+y^2=1\Longrightarrow$$
$$\left(\sqrt{1-y^2}\right)^2+\left(\sqrt{1-x^2}\right)^2=1\Longleftrightarrow$$
$$1-y^2+1-x^2=1\Longleftrightarrow$$
$$1-y^2+1-\left(\sqrt{1-y^2}\right)^2=1\Longleftrightarrow$$
$$1-y^2+1-(1-y^2)=1\Longleftrightarrow$$
$$1-y^2-(1-y^2)+1=1\Longleftrightarrow$$
$$0+1=1\Longleftrightarrow$$
$$1=1$$
A: As $\arccos  u+\arcsin u=\dfrac\pi2 $
from the given relation, $\arccos x+\arccos y=\dfrac\pi2$
Taking cosine in both sides, $$xy-\sqrt{(1-x^2)(1-y^2)}=0\iff xy=\sqrt{(1-x^2)(1-y^2)}$$
Now square both sides.
A: Using the identity $arcsiny + arccosy = \frac{\pi}{2}$
$$arcsinx + arcsiny=\frac{\pi}{2}$$
$$arcsinx + \frac{\pi}{2} -arcccosy=\frac{\pi}{2}$$
$$arcsinx =arccosy$$
Let $arcsinx = \theta =arccosy \iff cos\theta=y \land sin(\theta) =x$.
Using the pythagorean identity to show that $cos^2\theta +sin^2\theta=1 \iff y^2 + x^2 =1$
