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Trigonometric Ratios Of Compounds — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSTrigonometric Ratios Of Compounds1 Mark
MCQ
If in a $\Delta\text{ABC},\tan\text{A}+\tan\text{B}+\tan\text{C=0,}$ then $\cot\text{A}\cot\text{B}\cot\text{C}=$
  • A
    $6$
  • B
    $1$
  • C
    $\frac{1}{6}$
  • ✓
    None of these
✓

Answer

Correct option: D.
None of these
ABC is a tringle.
$\therefore\text{A}+\text{B}+\text{C}=\pi$
$\Rightarrow\text{A}+\text{B}+\pi-\text{C}$
$\Rightarrow\tan(\text{A}+\text{B})=\tan(\pi-\text{C})$
$\Rightarrow\frac{\tan\text{A}+\tan\text{B}}{1-\tan\text{A}\tan\text{B}}=-\tan\text{C}$
$\Rightarrow\tan\text{A}+\tan\text{B}=-\tan\text{C}+\tan\text{A}\tan\text{B}\tan\text{C}$
$\Rightarrow\tan\text{A}+\tan\text{B}+\tan\text{C}=\tan\text{A}\tan\text{B}\tan\text{C}$
$\Rightarrow0=\tan\text{A}+\tan\text{B}\tan\text{C}$ $\big[$ given: $\tan\text{A}\tan\text{B}\tan\text{C}=0\big]$
$\Rightarrow\tan\text{A}\tan\text{B}\tan\text{C}=0$
$\Rightarrow\frac{1}{\tan\text{A}\tan\text{B}\tan\text{C}}=\frac{1}{0}$
$\Rightarrow\cot\text{A}\cot\text{B}\cot\text{C}\rightarrow\infty$

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