If x, y , then the determinant = &- &1 & &1 (x+y)&- (x+y)&0 lies … - Vidyadip

MCQ

If $\text{x},\text{ y}\in\text{R},$ then the determinant $\triangle=\begin{vmatrix}\cos\text{x}&-\sin\text{x}&1\\\sin\text{x}&\cos\text{x}&1\\\cos(\text{x}+\text{y})&-\sin(\text{x}+\text{y})&0\end{vmatrix}$ lies in the interval:

✓
$\Big[-\sqrt{2},\sqrt{2}\Big]$
B
$[-1,1]$
C
$\Big[-\sqrt{2},1\Big]$
D
$\Big[-1,-\sqrt{2}\Big]$

✓

Answer

Correct option: A.

$\Big[-\sqrt{2},\sqrt{2}\Big]$

$\triangle=\begin{vmatrix}\cos\text{x}&-\sin\text{x}&1\\\sin\text{x}&\cos\text{x}&1\\\cos(\text{x}+\text{y})&-\sin(\text{x}+\text{y})&0\end{vmatrix}$
$=\begin{vmatrix}\cos\text{x}&-\sin\text{x}&1\\\sin\text{x}&\cos\text{x}&1\\0&0&\sin\text{y}-\cos\text{y}\end{vmatrix} [$Applying $R_3 \rightarrow R_3 - \text{cosy} R_1 + \text{siny} R_2]$
$=(\sin\text{y}-\cos\text{y})(\cos^2\text{x}+\sin^2\text{x})$
$=\sin\text{y}-\cos\text{y}$
$=\sqrt{2}\Big(\frac{1}{\sqrt{2}}\sin\text{y}-\frac{1}{\sqrt{2}}\cos\text{y}\Big)$
$=\sqrt{2}\Big(\cos\frac{\pi}{4}\sin\text{y}-\sin\frac{\pi}{4}\cos\text{y}\Big)$
$=\sqrt{2}\sin\Big(\text{y}-\frac{\pi}{4}\Big)$
Therefore, $-\sqrt{2}\leq\triangle\leq\sqrt{2}$
Hence, the correct option is $(a)$

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