JAMIA 2022 — Mathematics PYQ

In any triangle $ABC$, by the Law of Sines:

Thus, $\sin A=\frac{a}{2R}$, $\sin B=\frac{b}{2R}$, and $\sin C=\frac{c}{2R}$.

Also, recall the double angle formula: $\sin 2A=2\sin A\cos A$.

2. Expanding the Determinant

Substitute $\sin 2A=2\sin A\cos A$ and similar terms for $B$ and $C$:

3. Applying Geometric Constraints

For a general triangle, we can test a specific case (like an equilateral triangle) to find the value quickly, as the property must hold for all triangles $ABC$.

Let $A=B=C=60^{\circ }$ ($\frac{\pi }{3}$):

• $\sin A=\sin B=\sin C=\frac{\sqrt{3}}{2}$

• $\sin 2A=\sin 2B=\sin 2C=\sin 120^{\circ }=\frac{\sqrt{3}}{2}$

Substitute these values into the determinant:

Since all rows (or columns) are identical, the value of the determinant is:

4. General Verification

In the general expansion, the terms involving product of sines and cosines in a triangle $A+B+C=\pi$ cancel out to zero.

Final Answer:

The value of the determinant is $0$.