Mathematics - Determinant Question with Solution | TestHub

Perform row operations $R_2\to R_2-R_1$ and $R_3\to R_3-R_1$:$\Delta (x)=\left|\begin{array}{ccc}1& \cos x& \sin x\\ 0& \cos (x+\alpha )-\cos x& \sin (x+\alpha )-\sin x\\ 0& \cos (x+\beta )-\cos x& \sin (x+\beta )-\sin x\end{array}\right|$ Expand along the first column:$\Delta (x)=(\cos (x+\alpha )-\cos x)(\sin (x+\beta )-\sin x)-(\sin (x+\alpha )-\sin x)(\cos (x+\beta )-\cos x)$ This expression is of the form$AD-BC$. Let $A_1=\cos (x+\alpha )$, $B_1=\sin (x+\alpha )$, $A_2=\cos (x+\beta )$, $B_2=\sin (x+\beta )$.$\Delta (x)=(A_1-\cos x)(B_2-\sin x)-(B_1-\sin x)(A_2-\cos x)$ Expanding this, we get:$=A_1{B}_2-A_1\sin x-B_2\cos x+\cos x\sin x-(B_1{A}_2-B_1\cos x-A_2\sin x+\sin x\cos x)$$=A_1{B}_2-B_1{A}_2-A_1\sin x-B_2\cos x+B_1\cos x+A_2\sin x$ Rearranging terms:$=(\cos (x+\alpha )\sin (x+\beta )-\sin (x+\alpha )\cos (x+\beta ))+\sin x(\cos (x+\beta )-\cos (x+\alpha ))+\cos x(\sin (x+\alpha )-\sin (x+\beta ))$ Using the identity$\sin (A-B)=\sin A\cos B-\cos A\sin B$:$=\sin ((x+\beta )-(x+\alpha ))+(\sin x\cos (x+\beta )-\cos x\sin (x+\beta ))-(\sin x\cos (x+\alpha )-\cos x\sin (x+\alpha ))$$=\sin (\beta -\alpha )+\sin (x-(x+\beta ))-\sin (x-(x+\alpha ))$$=\sin (\beta -\alpha )+\sin (-\beta )-\sin (-\alpha )$$=\sin (\beta -\alpha )-\sin \beta +\sin \alpha$. This value is independent of $x$. So,$\Delta (x)=\sin \alpha -\sin \beta +\sin (\alpha -\beta )$.