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Q.
If $x , y \in R$, then the determinant
$\begin{vmatrix}\cos x & -\sin x & 1 \\ \sin x & \cos x & 1 \\ \cos (x+y) & -\sin (x+y) & 0\end{vmatrix}|$ 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, $-\sqrt{2} \leq \Delta \leq \sqrt{2}$