Question

For $\alpha ,\beta ,\gamma ,\theta ϵR$. Let
$A_{\theta }\left ( \alpha , \beta , \gamma \right )=\begin{vmatrix}
\cos \left ( \alpha +\theta \right ) & \sin \left ( \alpha +\theta \right ) & 1\\
\cos \left ( \beta +\theta \right ) & \sin \left ( \beta +\theta \right ) & 1\\
\cos \left ( \gamma +\theta \right ) & \sin \left ( \gamma +\theta \right ) & 1
\end{vmatrix}$

$\frac{d{A}_{\theta }}{d\theta }$ when $\theta =\frac{\pi }{6}$ equals

• A. $-1$
• B. $0$
• C. $1$
• D. none of these

Answer 1

The correct option is B $0$

$A_{\theta }(\alpha ,\beta ,\gamma )=\left|\begin{array}{ccc}\cos (\alpha +\theta )& \sin (\alpha +\theta )& 1\\ \cos (\beta +\theta )& \sin (\beta +\theta )& 1\\ \cos (\gamma +\theta )& \sin (\gamma +\theta )& 1\end{array}\right|$

$=\cos (\alpha +\theta )(\sin (\beta +\theta )-\sin (\gamma +\theta ))-\sin (\alpha +\theta )(\cos (\beta +\theta )-\cos (\gamma +\theta ))+(\cos (\beta +\theta )\sin (\gamma +\theta )-\sin (\beta +\theta )\cos (\gamma +\theta ))$

$=\sin (\beta -\gamma )+\sin (\alpha -\gamma )+\sin (\gamma -\beta )$

$\therefore A_{\theta }(\alpha ,\beta ,\gamma )=\sin (\beta -\gamma )+\sin (\alpha -\gamma )+\sin (\gamma -\beta )$

independent of $\theta$

$\therefore \frac{d{A}_{\theta }}{d\theta }=0$

Hence, option B.