UPSC NDA I 2024 Mathematics Paper

We can see that the given triangle is a right triangle because $65^2=16^2+63^2$.Therefore, angle $C$ is a right angle. Now we use the following trigonometric identities:$\cos 2A = 2\cos^2 A - 1$$\cos 2B = 2\cos^2 B - 1$$\cos 2C = 2\cos^2 C - 1$Now we know that, $\cos C = \cos 90^{\circ} = 0$.Therefore, $\cos 2C = 2\cos^2 C - 1 = 2 \times 0^2 - 1 = -1$.Hence, we have $\cos 2A + \cos 2B + \cos 2C = (2\cos^2 A - 1) + (2\cos^2 B - 1) + (-1) = 2(\cos^2 A + \cos^2 B) - 3$.Now, we use the identity $\cos^2 A + \cos^2 B = 1 + \cos A \cos B$ (which can be derived by using the angle subtraction formula for cosine: $\cos(A-B) = \cos A \cos B + \sin A \sin B$ and then squaring both sides and using the identity $\sin^2 x + \cos^2 x = 1$ ).Therefore, $\cos 2A + \cos 2B + \cos 2C = 2(1 + \cos A \cos B) - 3 = 2\cos A \cos B - 1$.Now, we can find $\cos A$ and $\cos B$ using the definition of cosine:$\cos A = \frac{AB}{AC} = \frac{16}{65} \text{ and } \cos B = \frac{BC}{AC} = \frac{63}{65} \text{. }$So, we get $\cos 2A + \cos 2B + \cos 2C = 2 \times \frac{16}{65} \times \frac{63}{65} - 1 = \frac{2016}{4225} - 1 = -1$.