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STD 11 - 3. trignometrical ratios,functions and identities — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsSTD 11 - 3. trignometrical ratios,functions and identities1 Mark
MCQ
If $A + B + C = {180^o},$ then the value of $\cot \frac{A}{2} + \cot \frac{B}{2} + \cot \frac{C}{2}$ will be
  • A
    $2\cot \frac{A}{2}\cot \frac{B}{2}\cot \frac{C}{2}$
  • B
    $4\cot \frac{A}{2}\cot \frac{B}{2}\cot \frac{C}{2}$
  • ✓
    $\cot \frac{A}{2}\cot \frac{B}{2}\cot \frac{C}{2}$
  • D
    $8\,\cot \frac{A}{2}\cot \frac{B}{2}\cot \frac{C}{2}$
✓

Answer

Correct option: C.
$\cot \frac{A}{2}\cot \frac{B}{2}\cot \frac{C}{2}$
c
(c) $A + B + C = {180^o}$, 

$\therefore \,\frac{A}{2} + \frac{B}{2} = {90^o} - \frac{C}{2}$

$\therefore \cot \left( {\frac{A}{2} + \frac{B}{2}} \right) = \cot \left( {{{90}^o} - \frac{C}{2}} \right)$

or $\frac{{\cot \frac{A}{2}\,.\,\cot \frac{B}{2}\, - 1}}{{\cot \frac{B}{2}\, + \,\cot \frac{A}{2}}} = \tan \frac{C}{2} = \frac{1}{{\cot \frac{C}{2}}}$ 

or $\left( {\cot \frac{A}{2}\cot \frac{B}{2} - 1} \right)\cot \frac{C}{2} = \cot \frac{B}{2} + \cot \frac{A}{2}$ 

$\cot \frac{A}{2}\,.\,\cot \frac{B}{2}\,.\,\cot \frac{C}{2} = \cot \frac{C}{2} + \cot \frac{B}{2}$ $ + \cot \frac{A}{2}.$

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