• # question_answer If A, B, C and D are the angles of a quadrilateral, then $\frac{\sum{\tan \,A}}{\sum{\cot \,A}}$ is equal to A)  $\Pi \,\,\tan \,\,A$                    B)  $\Pi \,\,\cot \,\,A$ C)  $\Sigma \,{{\tan }^{2}}A$             D)  $\Sigma \,{{\cot }^{2}}A$

$\because$   $A+B+C+D=2\pi$ or         $\tan \,(A+B+C+D)=0$ or $\frac{\Sigma \,\tan \,A-\Sigma \tan \,A\,\tan \,B\,\tan \,C}{1-\Sigma \tan \,A\,\tan \,B+\tan \,A\,\tan \,B\,\tan \,C\,D}=0$ $\Rightarrow$ $\Sigma \tan \,A-\Sigma \tan \,A\,\tan \,B\,\tan \,C=0$ $\Rightarrow$ $\Sigma \tan \,A=\tan \,A\,\tan \,B\,\tan \,C\,\tan D\,\,\Sigma \cot \,A$ $\Rightarrow$ $\frac{\Sigma \,\tan \,A}{\Sigma \cot \,A}=\Pi \,\tan \,A$