MCQ
If in $\text{a}\triangle\text{ABC},\tan\text{B}+\tan\text{C}=6,$ then $\cot\text{A}\cot\text{B}\cot\text{C}=$
  • A
    $6$
  • B
    $1$
  • $\frac16$
  • D
    None of these

Answer

Correct option: C.
$\frac16$
In $\triangle\text{ABC},$
$\text{A+B+C}=\pi$
We know that $\tan(\text{A+B+C)}=\frac{\tan\text{A}+\tan\text{B}+\tan\text{C}-\tan\text{A}\tan\text{B}\tan\text{C}}{1-\tan\text{A}\tan\text{B}-\tan\text{B}\tan\text{C}-\tan\text{C}\tan\text{A}}$ and $\tan\pi=0.$
$\therefore\tan\text{A}+\tan\text{B}+\tan\text{C}-\tan\text{A}\tan\text{B}\tan\text{C}=0$
$\tan\text{A}+\tan\text{B}+\tan\text{C}=\tan\text{A}\tan\text{B}\tan\text{C}$
If $\tan\text{A}+\tan\text{B}+\tan\text{C}=6,\tan\text{A}\tan\text{B}\tan\text{C}=6$
$\Rightarrow\frac{1}{\tan\text{A}\tan\text{B}\tan\text{C}}=\frac16$
$\Rightarrow\cot\text{A}\cot\text{B}\cot\text{C}=\frac16$

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