Question

${A_{\theta }}^2+{A_{\varphi }}^2-2{\left(A_{\theta +\varphi }\right)}^2$ equals

• A. $-2A_{\theta }{A}_{\varphi }$
• B. $A_{\theta }+A_{\varphi }$
• C. $A_{\theta }-A_{\varphi }$
• D. none of these

Answer 1

The correct option is C $A_{\theta }-A_{\varphi }$

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

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

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

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

$A_{\theta }$ is independent of $\theta$

Similarly,

$A_{\varphi }$ and $A_{\theta +\varphi }$ are independent of $\varphi$ and $\theta +\varphi$ respectively.

$\therefore {A_{\theta }}^2+{A_{\varphi }}^2-2{\left(A_{\theta +\varphi }\right)}^2=0$

$A_{\theta }-A_{\varphi }=0$

Hence, option C.