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DETERMINANTS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDETERMINANTS5 Marks
Question
If A + B + C = 0, then prove that $\begin{vmatrix}1&\cos\text{C}&\cos\text{B}\\\cos\text{C}&1&\cos\text{A}\\\cos\text{B}&\cos\text{A}&1\end{vmatrix}=0.$
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Answer

$\text{L.H.S}=\begin{vmatrix}1&\cos\text{C}&\cos\text{B}\\\cos\text{C}&1&\cos\text{A}\\\cos\text{B}&\cos\text{A}&1\end{vmatrix}$
$=(1-\cos^2\text{A})-\cos\text{C}(\cos\text{C}-\cos\text{A}.\cos\text{B})+\cos\text{B}(\cos\text{C}.\cos\text{A}-\cos\text{B})$
$=\sin^2\text{A}-\cos^2\text{C}+\cos\text{A}.\cos\text{B}.\cos\text{C}+\cos\text{A}.\cos\text{B}.\cos\text{C}-\cos^2\text{B}$
$=\sin^2\text{A}-\cos^2\text{B}+2\cos\text{A}.\cos\text{B}.\cos\text{C}-\cos^2\text{C}$
$=-\cos(\text{A}+\text{B}).\cos(\text{A}-\text{B})+2\cos\text{A}.\cos\text{B}.\cos\text{C}-\cos^2\text{C}$ $\big[\because\ \cos^2\text{B}-\sin^2\text{A}=\cos(\text{A}+\text{B}).\cos(\text{A}-\text{B}).\big]$
$=-\cos(-\text{C}).\cos(\text{A}-\text{B})+\cos\text{C}(2\cos\text{A}.\cos\text{B}-\cos\text{C})$ $[\because\ \cos(-\theta)=\cos\theta]$
$=-\cos\text{C}.(\cos\text{A}.\cos\text{B}+\sin\text{A}.\sin\text{B}-2\cos\text{A}.\cos\text{B}-\cos\text{C})$
$=\cos\text{C}(\cos\text{A}.\cos\text{B}-\sin\text{A}.\sin\text{B}-\cos\text{C})$
$=\cos\text{C}[\cos(\text{A}+\text{B})-\cos\text{C}]$
$=\cos\text{C}(\cos\text{C}-\cos\text{C})=0$
$=\text{R.H.S}$
Hence proved.

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