Using properties of determinants, prove that:
$\left| {\begin{array}{*{20}{l}}
{\sin \alpha }&{\cos \alpha }&{\cos (\alpha + \delta )} \\
{\sin \beta }&{\cos \beta }&{\cos (\beta + \delta )} \\
{\sin \gamma }&{\cos \gamma }&{\cos (\gamma + \delta )}
\end{array}} \right| = 0$
Using properties of determinants, prove that:
$\left| {\begin{array}{*{20}{l}}
{\sin \alpha }&{\cos \alpha }&{\cos (\alpha + \delta )} \\
{\sin \beta }&{\cos \beta }&{\cos (\beta + \delta )} \\
{\sin \gamma }&{\cos \gamma }&{\cos (\gamma + \delta )}
\end{array}} \right| = 0$
$\Delta=\left|\begin{array}{lll}\sin \alpha & \cos \alpha & \cos (\alpha+\delta) \\ \sin \beta & \cos \beta & \cos (\beta+\delta) \\ \sin \gamma & \cos \gamma & \cos (\gamma+\delta)\end{array}\right|$
$=\frac{1}{\sin \delta \cos \delta}\left|\begin{array}{ccc}\sin \alpha \sin \delta & \cos \alpha \cos \delta & \cos \alpha \cos \delta-\sin \alpha \sin \delta \\ \sin \beta \sin \delta & \cos \beta \cos \delta & \cos \beta \cos \delta-\sin \beta \sin \delta \\ \sin \gamma \sin \delta & \cos \gamma \cos \delta & \cos \gamma \cos \delta-\sin \gamma \sin \delta\end{array}\right|$
Applying $C_{1} \rightarrow+C_{1}+C_{3},$ we have:
$\Delta=\frac{1}{\sin \delta \cos \delta}\left|\begin{array}{lll}
\cos \alpha \cos \delta & \cos \alpha \cos \delta & \cos \alpha \cos \delta-\sin \alpha \sin \delta \\
\cos \beta \cos \delta & \cos \beta \cos \delta & \cos \beta \cos \delta-\sin \beta \sin \delta \\
\cos \gamma \cos \delta & \cos \gamma \cos \delta & \cos \gamma \cos \delta-\sin \gamma \sin \delta
\end{array}\right|$
Here, two columns $C_{1}$ and $C_{2}$ are identical.
$\therefore \Delta=0$
Hence, the given result is proved.
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If $\mathrm{a, b, c}$ are positive and unequal, show that value of the determinant $\Delta=\left|\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right|$ is negative.
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Using the property of determinants and without expanding, prove that:
$\left|\begin{array}{lll}b+c & q+r & y+z \\ c+a & r+p & z+x \\ a+b & p+q & x+y\end{array}\right|=2\left|\begin{array}{lll}a & p & x \\ b & q & y \\ c & r & z\end{array}\right|$
Using the property of determinants and without expanding, prove that:
$\left|\begin{array}{lll}b+c & q+r & y+z \\ c+a & r+p & z+x \\ a+b & p+q & x+y\end{array}\right|=2\left|\begin{array}{lll}a & p & x \\ b & q & y \\ c & r & z\end{array}\right|$
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If $x, y, z$ are different and $\Delta=\left|\begin{array}{lll}x & x^{2} & 1+x^{2} \\ y & y^{2} & 1+y^{2} \\ z & z^{2} & 1+z^{2}\end{array}\right|=0,$ then show that $1+x y z=0$.