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Determinants — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDeterminants1 Mark
MCQ
If $\text{A}+\text{B}+\text{C}=\pi,$ then the value of $\begin{vmatrix}\sin(\text{A}+\text{B}+\text{C})&\sin(\text{A}+\text{C})&\cos\text{C}\\-\sin\text{B}&0&\tan\text{A}\\\cos(\text{A}+\text{B})&\tan(\text{B}+\text{C})&0\end{vmatrix}$ is equal to:
  • ✓
    $0$
  • B
    $1$
  • C
    $2\sin\text{B}\tan\text{A}\cos\text{C}$
  • D
    None of these.
✓

Answer

Correct option: A.
$0$
$\text{A}+\text{B}+\text{C}=\pi$
$\text{A}+\text{C}=\pi-\text{B},\text{A}+\text{B}=\pi-\text{C}$ and $\text{B}+\text{C}=\pi-\text{A}$
Thus the determinant becomes
$\begin{vmatrix}\sin\pi&\sin(\pi-\text{B})&\cos\text{C}\\-\sin\text{B}&0&\tan\text{A}\\\cos(\pi-\text{C})&\tan(\pi-\text{A})&0\end{vmatrix}$
$=\begin{vmatrix}0&\sin\text{B}&\cos\text{C}\\-\sin\text{B}&0&\tan\text{A}\\-\cos\text{C}&-\tan\text{A}&0\end{vmatrix}$
$[\sin\pi=0,\sin(\pi-\text{B}),\cos(\pi-\text{C})=-\cos\text{C},\tan(\pi-\text{A})=-\tan\text{A}]$
It is a skew symmetric matrix of the odd order $3.$
Thus by property of determinants, we get
$|\triangle|=0$
$\begin{vmatrix}0&\sin\text{B}&\cos\text{C}\\-\sin\text{B}&0&\tan\text{A}\\-\cos\text{C}&-\tan\text{A}&0\end{vmatrix}$

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