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  • If A, B and C are angles of a triangle, then the determinant is equal to A. 0

If A, B and C are angles of a triangle, then the determinant \begin{array}{|ccc|} -1 & \cos \mathrm{C} & \cos \mathrm{B} \\ \cos \mathrm{C} & -1 & \cos \mathrm{A} \\ \cos \mathrm{B} & \cos \mathrm{A} & -1 \end{array} \mid is equal to
A. 0

B. –1
C. 1
D. None of these

Answers (1)

A)

Given:

\begin{array}{|ccc|} -1 & \cos \mathrm{C} & \cos \mathrm{B} \\ \cos \mathrm{C} & -1 & \cos \mathrm{A} \\ \cos \mathrm{B} & \cos \mathrm{A} & -1 \end{array} \mid

Expand along Column 1

\begin{array}{r} |\mathrm{A}|=\mathrm{a}_{11}(-1)^{1+1}\left|\begin{array}{ll} \mathrm{a}_{22} & \mathrm{a}_{23} \\ \mathrm{a}_{32} & \mathrm{a}_{33} \end{array}\right|+\mathrm{a}_{21}(-1)^{2+1}\left|\begin{array}{ll} \mathrm{a}_{12} & \mathrm{a}_{13} \\ \mathrm{a}_{32} & \mathrm{a}_{33} \end{array}\right| +\mathrm{a}_{31}(-1)^{3+1}\left|\begin{array}{ll} \mathrm{a}_{12} & \mathrm{a}_{13} \\ \mathrm{a}_{22} & \mathrm{a}_{23} \end{array}\right| \end{array}

\Delta=(-1)\left|\begin{array}{cc} -1 & \cos A \\ \cos A & -1 \end{array}\right|-\cos C\left|\begin{array}{cc} \cos C & \cos B \\ \cos A & -1 \end{array}\right|+\cos B\left|\begin{array}{cc} \mid \cos C & \cos B \\ -1 & \cos A \end{array}\right|

\\ \vspace{\baselineskip}= [(-1)$ \{ $ 1 -cos\textsuperscript{2}A$ \} $ -cos C$ \{ $ -cos C -cos Acos B$ \} $ + cos B$ \{ $ cos A cos C + cos B$ \} $ ]\\ \\ \vspace{\baselineskip}= -1 + cos\textsuperscript{2}A + cos\textsuperscript{2}C + cos A cos B cos C + cos A cos B cos C + cos\textsuperscript{2}B\\ \\ \vspace{\baselineskip}= -1 + cos\textsuperscript{2}A + cos\textsuperscript{2}B + cos\textsuperscript{2}C + 2cos A cos B cos C\\ \vspace{\baselineskip} \text{Using the formula}\\ \vspace{\baselineskip} 1 + cos2A = 2cos\textsuperscript{2}A\\

\\ \begin{aligned} &=-1+\frac{1+\cos 2 \mathrm{~A}}{2}+\frac{1+\cos 2 \mathrm{~B}}{2}+\frac{1+\cos 2 \mathrm{C}}{2}+2 \cos \mathrm{A} \cos \mathrm{B} \cos \mathrm{C}\\ &\text { Taking L.C.M, we get }\\ &=\frac{-2+1+\cos 2 A+1+\cos 2 B+1+\cos 2 C+4 \cos A \cos B \cos C}{2}\\ &=\frac{1+(\cos 2 A+\cos 2 B)+\cos 2 C+4 \cos C \cos A \cos B}{2}\\ &\text { Now use: } \cos (A+B) \cos (A-B)=2 \cos A \cos B\\ &so, \cos 2 A+\cos 2 B=2 \cos (A+B) \cos (A-B) \end{aligned}

 

\\ \begin{aligned} &=\frac{1+\cos 2 C+\{2 \cos (A+B) \cos (A-B)\}+4 \cos A \cos B \cos C}{2}\\ &=\frac{1+2 \cos ^{2} C-1+\{2 \cos (A+B) \cos (A-B)\}+4 \cos A \cos B \cos C}{2}\\ &.=\frac{2 \cos ^{2} \mathrm{C}+[2 \cos (\mathrm{A}+\mathrm{B}) \cos (\mathrm{A}-\mathrm{B})\}+4 \cos \mathrm{A} \cos \mathrm{B} \cos C}{2} \ldots (\mathrm{i})\\ &\text { We know that } A, B, C \text { are angles of triangle }\\ &\Rightarrow A+B+C=\pi\\ &\Rightarrow A+B=\pi-C \end{aligned}

\\ =\frac{2 \cos ^{2} C+\{2 \cos (\pi-C) \cos (A-B)\}+4 \cos A \cos B \cos C}{2} \\ =\frac{2 \cos ^{2} C+\{-2 \cos C \cos (A-B)\}+4 \cos A \cos B \cos C}{2} [\because \cos (\pi-x)=-\cos x] \\ =\frac{-2 \cos C\{\cos (A-B)-\cos C\}+4 \cos A \cos B \cos C}{2}

\\ \vspace{\baselineskip}= -cos C$ \{ $ cos(A -B) -cos C$ \} $ + 2cos Acos Bcos C\\ \\ \vspace{\baselineskip}= -cos C[cos(A -B) -cos$ \{ $ $ \pi $ -(A + B)$ \} $ ] + 2cos Acos Bcos C\\ \\ \vspace{\baselineskip}= -cos C[cos(A -B) + cos(A + B)] + + 2cos Acos Bcos C\\ \\ \vspace{\baselineskip}= -cos C[2cos Acos B] + 2cos Acos Bcos C\\ \\ \vspace{\baselineskip}= 0\\

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