Search any question & find its solution
Question:
Answered & Verified by Expert
If $A+B+C=0$, then prove that $\left|\begin{array}{ccc}1 & \cos C & \cos B \\ \cos C & 1 & \cos A \\ \cos B & \cos A & 1\end{array}\right|=0$.
Solution:
1944 Upvotes
Verified Answer
$\because \cos (\mathrm{A}+\mathrm{B})=\cos (-\mathrm{C})$
$\therefore \mathrm{LHS}=\left|\begin{array}{ccc}1 & \cos C & \cos \mathrm{B} \\ \cos \mathrm{C} & 1 & \cos \mathrm{A} \\ \cos \mathrm{B} & \cos \mathrm{A} & 1\end{array}\right|$
$=1\left(1-\cos ^2 A\right)-\cos C(\cos C-\cos A \cdot \cos B)+\cos B$
$(\cos C \cdot \cos A-\cos B)$
$=\sin ^2 \mathrm{~A}-\cos ^2 \mathrm{~B}+2 \cos \mathrm{A} \cdot \cos \mathrm{B} \cdot \cos \mathrm{C}-\cos ^2 \mathrm{C}$
$=-\cos (\mathrm{A}+\mathrm{B}) \cdot \cos (\mathrm{A}-\mathrm{B})+2 \cos \mathrm{A} \cdot \cos \mathrm{B}$
$\times \cos C-\cos ^2 \mathrm{C}$
$\left[\because \cos ^2 \mathrm{~B}-\sin ^2 \mathrm{~A}=\cos (\mathrm{A}+\mathrm{B}) \cdot \cos (\mathrm{A}-\mathrm{B})\right]$
$=-\cos (-\mathrm{C}) \cdot \cos (\mathrm{A}-\mathrm{B})+\cos \mathrm{C}(2 \cos \mathrm{A} \cdot \cos \mathrm{B}-\cos \mathrm{C})$
$[\because \cos (-\theta)=\cos \theta]$
$=-\cos C(\cos A \cdot \cos \mathrm{B}+\sin \mathrm{A} \cdot \sin \mathrm{B}$
$-2 \cos A \cdot \cos B+\cos C$ )
$=\cos \mathrm{C}(\cos \mathrm{A} \cdot \cos \mathrm{B}-\sin \mathrm{A} \cdot \sin \mathrm{B}-\cos \mathrm{C})$
$=\cos C[\cos (A+B)-\cos C]$
$=\cos C(\cos C-\cos C)=0=$ RHS
Hence proved.
$\therefore \mathrm{LHS}=\left|\begin{array}{ccc}1 & \cos C & \cos \mathrm{B} \\ \cos \mathrm{C} & 1 & \cos \mathrm{A} \\ \cos \mathrm{B} & \cos \mathrm{A} & 1\end{array}\right|$
$=1\left(1-\cos ^2 A\right)-\cos C(\cos C-\cos A \cdot \cos B)+\cos B$
$(\cos C \cdot \cos A-\cos B)$
$=\sin ^2 \mathrm{~A}-\cos ^2 \mathrm{~B}+2 \cos \mathrm{A} \cdot \cos \mathrm{B} \cdot \cos \mathrm{C}-\cos ^2 \mathrm{C}$
$=-\cos (\mathrm{A}+\mathrm{B}) \cdot \cos (\mathrm{A}-\mathrm{B})+2 \cos \mathrm{A} \cdot \cos \mathrm{B}$
$\times \cos C-\cos ^2 \mathrm{C}$
$\left[\because \cos ^2 \mathrm{~B}-\sin ^2 \mathrm{~A}=\cos (\mathrm{A}+\mathrm{B}) \cdot \cos (\mathrm{A}-\mathrm{B})\right]$
$=-\cos (-\mathrm{C}) \cdot \cos (\mathrm{A}-\mathrm{B})+\cos \mathrm{C}(2 \cos \mathrm{A} \cdot \cos \mathrm{B}-\cos \mathrm{C})$
$[\because \cos (-\theta)=\cos \theta]$
$=-\cos C(\cos A \cdot \cos \mathrm{B}+\sin \mathrm{A} \cdot \sin \mathrm{B}$
$-2 \cos A \cdot \cos B+\cos C$ )
$=\cos \mathrm{C}(\cos \mathrm{A} \cdot \cos \mathrm{B}-\sin \mathrm{A} \cdot \sin \mathrm{B}-\cos \mathrm{C})$
$=\cos C[\cos (A+B)-\cos C]$
$=\cos C(\cos C-\cos C)=0=$ RHS
Hence proved.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.