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Trigonometric Functions — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsTrigonometric Functions2 Marks
MCQ
In a $\triangle A B C$, if $b ^2+ c ^2=3 a ^2$, then $\cot B+$ $\cot C$ $-\cot A=$
  • A
    $1$
  • B
    $\frac{ ab }{4 \Delta}$
  • ✓
    $0$
  • D
    $\frac{\text { ac }}{4 \Delta}$
✓

Answer

Correct option: C.
$0$
(C) $\cot B +\cot C -\cot A =\frac{\cos B }{\sin B }+\frac{\cos C }{\sin C }-\cot A$
$=\frac{\sin C \cos B +\cos C \sin B }{\sin B \sin C }-\cot A$
$=\frac{\sin ( B + C )}{\sin B \sin C }-\frac{\cos A }{\sin A }$
$=\frac{\sin ^2 A-\sin B \sin C \cos A}{\sin A \sin B \sin C}=\frac{a^2-b c \cos A}{(a b c)}$
$=\frac{ a ^2- bc \frac{\left( b ^2+ c ^2- a ^2\right)}{2 bc }}{( abc )}$
$=\frac{3 a ^2- b ^2- c ^2}{2( abc )}=\frac{3 a ^2-\left( b ^2+ c ^2\right)}{2( abc )}$
$\therefore \quad \cot B +\cot C -\cot A =\frac{3 a ^2-3 a ^2}{2( abc )}=0$ $\ldots\left[\because b^2+c^2=3 a^2\right]$

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