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DETERMINANTS — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDETERMINANTS3 Marks
Question
Using properties of determinants, prove that: $\begin{vmatrix}\sin\alpha&\cos\alpha&\cos(\alpha+\delta)\\\sin\beta&\cos\beta&\cos(\beta+\delta)\\\sin\gamma&\cos\gamma&\cos(\gamma+\delta)\end{vmatrix}=0$
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Answer

$\triangle=\begin{vmatrix}\sin\alpha&\cos\alpha&\cos(\alpha+\delta)\\\sin\beta&\cos\beta&\cos(\beta+\delta)\\\sin\gamma&\cos\gamma&\cos(\gamma+\delta)\end{vmatrix}$
$=\frac{1}{\sin\delta\cos\delta}\begin{vmatrix}\sin\alpha\sin\delta&\cos\alpha\cos\delta&\cos\alpha\cos\delta-\sin\alpha\sin\delta\\\sin\beta\sin\delta&\cos\beta\cos\delta&\cos\beta\cos\delta-\sin\beta\sin\delta\\\sin\gamma\sin\delta&\cos\gamma\cos\delta&\cos\gamma\cos\delta-\sin\gamma\sin\delta\end{vmatrix}$
Applying $C_1 \rightarrow C_1 + C_3,$ we have$:$
$\triangle=\frac{1}{\sin\delta\cos\delta}\begin{vmatrix}\cos\alpha\cos\delta&\cos\alpha\cos\delta&\cos\alpha\cos\delta-\sin\alpha\sin\delta\\\cos\beta\cos\delta&\cos\beta\cos\delta&\cos\beta\cos\delta-\sin\beta\sin\delta\\\cos\gamma\cos\delta&\cos\gamma\cos\delta&\cos\gamma\cos\delta-\sin\gamma\sin\delta\end{vmatrix}$
Here, two columns $C_1$ and $C_1 + C_3,$ are identical.
$\therefore\triangle=0.$
Hence, the given result is proved.

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