The value of \( \Delta=\begin{vmatrix}2\sin 40^\circ & 2\sin 50^\circ\\-3\cos 40^\circ & 3\cos 50^\circ\end{vmatrix} \) is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 6
Expanding the determinant:
\( \displaystyle \Delta=(2\sin 40^\circ)(3\cos 50^\circ)-(2\sin 50^\circ)(-3\cos 40^\circ) \)
\( \displaystyle =6\left(\sin 40^\circ\cos 50^\circ+\cos 40^\circ\sin 50^\circ\right) \)
Using \( \sin A\cos B+\cos A\sin B=\sin(A+B) \):
\( \displaystyle \Delta=6\sin(40^\circ+50^\circ) \)
\( \displaystyle =6\sin 90^\circ=6 \)
Shortcut: Since 40° + 50° = 90°, \( \sin 50^\circ=\cos 40^\circ \) and \( \cos 50^\circ=\sin 40^\circ \) (complementary angles):
\( \displaystyle \Delta=6\sin^2 40^\circ+6\cos^2 40^\circ=6 \)