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The value of Δ=∣2sin⁡40∘2sin⁡50∘−3cos⁡40∘3cos⁡50∘∣\Delta = \begin{vmatrix} 2\sin40^\circ & 2\sin50^\circ \\ -3\cos40^\circ & 3\cos50^\circ \end{vmatrix} is:

Solution

✅ Correct Option: 4

Δ=6sin⁡40∘cos⁡50∘+6sin⁡50∘cos⁡40∘\Delta = 6\sin40^\circ\cos50^\circ + 6\sin50^\circ\cos40^\circ.

=6sin⁡(40∘+50∘)=6sin⁡90∘=6=6\sin(40^\circ+50^\circ)=6\sin90^\circ=6.

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