Question

Determinant $\left|\begin{array}{ccc}1 & x & y \\ 2 & \sin x+2 x & \sin y+2 y \\ 3 & \cos x+3 x & \cos y+3 y\end{array}\right|$ is equal to

(A) $\sin (x-y)$

(B) $\cos (x-y)$

(C) $\cos (x+y)$

(D) $x y \sin (x-y)$

Answer 1

Applying $R_2 \rightarrow R_2-2 R_1, R_3 \rightarrow R_3-3 R_1$

$\begin{aligned}\Delta & =\left|\begin{array}{ccc}1 & x & y \\0 & \sin x & \sin y \\0 & \cos x & \cos y\end{array}\right| \\& =\sin x \cos y-\cos x \sin y=\sin (x-y)\end{aligned}$

So, The correct option is (A).