Animated Solution for Mathematics - Matrices and Determinants: If ${\Delta }_1=\left|\begin{array}{ccc}x& \sin \theta & \cos \theta \\ -\sin \theta & -x& 1\\ \cos \theta & 1& x\end{array}\right|$ and ${\Delta }_2=\left|\begin{array}{ccc}x& \sin 2\theta & \cos 2\theta \\ -\sin 2\theta & -x& 1\\ \cos 2\theta & 1& x\end{array}\right|$, $x\ne 0$; then for all $\theta \in (0,\frac{\pi }{2})$ :

Imagine you are standing before a complex-looking matrix filled with $x$, $\sin \theta$, and $\cos \theta$. At first glance, it looks like a chaotic mess of variables.

In the world of JEE Advanced, chaos is often just order in disguise. Today, we are going to peel back the layers of this problem by observing the elegant geometry of algebra.

We begin with ${\Delta }_1$. Our weapon of choice is the cofactor expansion along the first row ($R_1$). Remember, the sign convention for a $3\times 3$ determinant is $+-+$.

We take the first element, $x$, and multiply it by the minor determinant: This gives us $x(-x^2-1)$, which simplifies to $-x^3-x$.

Now, we move to the second element, $\sin \theta$. We must attach a negative sign, giving us: Expanding this, we get $-\sin \theta (-x\sin \theta -\cos \theta )$, which simplifies to $x{\sin }^2\theta +\sin \theta \cos \theta$.

Finally, the third element, $\cos \theta$, gives us: This results in $\cos \theta (-\sin \theta +x\cos \theta )$, or $-\sin \theta \cos \theta +x{\cos }^2\theta$.

Now, let us bring these pieces together:

Look closely at the terms. The term $+\sin \theta \cos \theta$ and $-\sin \theta \cos \theta$ are staring at each other, waiting to vanish. They cancel out perfectly!

This is the moment where the complexity begins to dissolve. We are left with:

We are almost there. Factor out the $x$ from the last two terms:

Here, the fundamental identity of trigonometry, ${\sin }^2\theta +{\cos }^2\theta =1$, comes to our rescue. Substituting this, we get:

The $-x$ and $+x$ cancel out, leaving us with the elegant result:

Now, look at ${\Delta }_2$. It is identical to ${\Delta }_1$, but with $2\theta$ instead of $\theta$. Does the angle matter? No!

Since ${\Delta }_1$ is independent of $\theta$, ${\Delta }_2$ must be independent of $2\theta$. Therefore, ${\Delta }_2=-x^3$.

Adding them together, we find:

We have conquered the problem not by fighting the trigonometry, but by letting the algebra reveal its own truth. Keep this mindset, and no determinant will ever intimidate you again.