Question

If $A + B + C = 180^\circ$, then the value of $\begin{bmatrix} 1& 1 & 0 \\\[0.3em] -\cos\,C & \cot\, A & -\cot\, A \\\[0.3em] -\cot\,B & \cot\,C & \cot\, B \end{bmatrix}$ is

• A. $1 + \cot A \cot B$
• B. $1 + \cot B \cot C$
• C. $1 + \cot C \cot A$
• D. none of these.

Answer 1

Answer: (A)

$1 + \cot A \cot B$

Answer 2

$\begin{vmatrix} 1& 1 & 0 \\\[0.3em] -\cos\,C & \cot\, A & -\cot\, A \\\[0.3em] -\cot\,B & \cot\,C & \cot\, B \end{vmatrix}$ = $\begin{vmatrix} 1& 1 & 0 \\\[0.3em] -\cos\,C & \cot\, A +\cot \, C & -\cot\, A \\\[0.3em] -\cot\,B & \cot \, B + \cot\,C & \cot\, B \end{vmatrix}$ = $\cot A \cot B + \cot B \cot C + \cot A \cot B + \cot A \cot C$ = $\cot A \cot B + (\cot A \cot B + \cot B \cot C + \cot C \cot A)$ Since $A + B + C = 108^\circ$ $\therefore\: \tan (A + B) = \tan (180^\circ - C) = - \tan C$ $\Rightarrow$ $\frac{\tan \,A \,\tan \,B }{1 - \tan \, A \, \tan \, B} = - \, \tan \, C$ $\Rightarrow \: \tan A + \tan B + \tan C = \tan A \tan B \tan C$ $\Rightarrow\:\: \frac{\tan \, A}{\tan\, A\, \tan\, B\, \tan \,C } + \frac{\tan \, B}{\tan\, A\, \tan\, B\, \tan \,C } +\frac {\tan \,C}{\tan\, A\, \tan\, B\, \tan \,C } = 1$ $\Rightarrow\: \cot B \cot C + \cot C \cot A + \cot A \cot B = 1$ $\therefore$ given value = $1 + \cot A \cot B$