Animated Solution for Mathematics - Inverse Trigonometric Functions: Let $(x,y)$ be such that ${\sin }^{-1}(ax)+{\cos }^{-1}(y)+{\cos }^{-1}(bxy)=\frac{\pi }{2}$. Match the statements in Column I with statements in Column II and indicate your answer by darkening the appropriate bubbles in the $4\times 4$ matrix given in the ORS.

Welcome, fellow explorer of the mathematical landscape. Today, we are going to dismantle a problem that, at first glance, looks like a tangled mess of inverse trigonometric functions.

In JEE Advanced, the most complex-looking problems often hide a beautiful, simple symmetry. Let us embark on this journey together.

The Power of the Complementary Identity

We start with the given equation:

When you see $\frac{\pi }{2}$ on the right-hand side, your mind should immediately race to the complementary angle identity: ${\sin }^{-1}(\theta )+{\cos }^{-1}(\theta )=\frac{\pi }{2}$. This is our golden key.

By rearranging our original equation to ${\cos }^{-1}(y)+{\cos }^{-1}(bxy)=\frac{\pi }{2}-{\sin }^{-1}(ax)$, we can instantly transform the right side into ${\cos }^{-1}(ax)$. Now, our equation is uniform:

This uniformity is the first step toward victory.

The Substitution Strategy

Now, we have three inverse cosine terms. Let us simplify our mental load.

Let $\alpha ={\cos }^{-1}(y)$, $\beta ={\cos }^{-1}(bxy)$, and $\gamma ={\cos }^{-1}(ax)$. Our equation is now simply $\alpha +\beta =\gamma$.

To get back to the algebraic world, we isolate $\beta$ and take the cosine of both sides: $\cos (\beta )=\cos (\gamma -\alpha )$.

Using the expansion formula $\cos (\gamma -\alpha )=\cos (\gamma )\cos (\alpha )+\sin (\gamma )\sin (\alpha )$, we translate our angles back into variables.

We know $\cos (\beta )=bxy$, $\cos (\gamma )=ax$, and $\cos (\alpha )=y$.

Recalling that $\sin (\theta )=\sqrt{1-{\cos }^2(\theta )}$, we find $\sin (\gamma )=\sqrt{1-a^2{x}^2}$ and $\sin (\alpha )=\sqrt{1-y^2}$.

The Radical Dance

Substituting these back, we get:

This looks intimidating, but do not panic. We isolate the radical term:

Now, we square both sides to eliminate the radicals. This gives us:

This is our master equation, the engine that will drive our solutions for all cases.

Unveiling the Geometry

Now, we test our specific cases. When $a=1$ and $b=0$, our equation becomes $(0-1{)}^2{x}^2{y}^2=(1-x^2)(1-y^2)$, which simplifies to $x^2{y}^2=1-x^2-y^2+x^2{y}^2$.

The $x^2{y}^2$ terms cancel out, leaving $x^2+y^2=1$, which is the unit circle.

When $a=1$ and $b=1$, the left side vanishes, leaving $(1-x^2)(1-y^2)=0$, which represents the lines $x^2=1$ or $y^2=1$.

Finally, when $a=2$ and $b=2$, we get $0=(1-4x^2)(1-y^2)$, leading to $4x^2=1$ or $y^2=1$.

Each case reveals a distinct geometric shape. You see, the math was not trying to confuse you; it was trying to show you the hidden geometry of the plane. Keep practicing this systematic approach, and you will find that even the most daunting problems are just puzzles waiting to be solved.