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Q.
If $x, y \in R$, then the determinant $\begin{vmatrix}\cos x & -\sin x & 1 \\ \sin x & \cos x & 1 \\ \cos (x+y) & -\sin (x+y) & 0\end{vmatrix}$ lie lies in the interval
Determinants
Solution:
The given determinant is
$\begin{vmatrix}\cos x & -\sin x & 1 \\ \sin x & \cos x & 1 \\ \cos (x+y) & -\sin (x+y) & 0\end{vmatrix}$
Applying $R_3 \rightarrow R_3-\cos y R_1+\sin y R_2$, we get
$\Delta=\begin{vmatrix}\cos x & -\sin x & 1 \\ \sin x & \cos x & 1 \\ 0 & 0 & \sin y-\cos y\end{vmatrix}$
By expanding along $R_3$, we have
$\Delta =(\sin y-\cos y)\left(\cos ^2 x+\sin ^2 x\right) $
$ =(\sin y-\cos y)=\sqrt{2}\left[\frac{1}{\sqrt{2}} \sin y-\frac{1}{\sqrt{2}} \cos y\right]$
$ =\sqrt{2}\left[\cos \frac{\pi}{4} \sin y-\sin \frac{\pi}{4} \cos y\right]=\sqrt{2} \sin \left(y-\frac{\pi}{4}\right)$
Hence, $-\sqrt{2} \leq \Delta \leq \sqrt{2}$