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)  \[\Sigma \,{{\cot }^{2}}A\]

Correct Answer: A

Solution :

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