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Trigonometric Functions — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsTrigonometric Functions2 Marks
MCQ
In $\triangle A B C,\left(\cot \frac{A}{2}+\cot \frac{B}{2}\right)\left(a \sin ^2 \frac{B}{2}+b \sin ^2 \frac{A}{2}\right)=$
  • A
    $\cot C$
  • B
    $c \cot C$
  • C
    $\cot \frac{ C }{2}$
  • ✓
    $c \cot \frac{C}{2}$
✓

Answer

Correct option: D.
$c \cot \frac{C}{2}$
(D) $\{\cot \frac{ A }{2}+\cot \frac{ B }{2}\}\{ asin ^2 \frac{B}{2}+ b \sin ^2 \frac{A}{2}\}$
$=\{\frac{\cos \frac{ C }{2}}{\sin \frac{A}{2} \sin \frac{B}{2}}\}\{a \sin ^2 \frac{B}{2}+ b \sin ^2 \frac{A}{2}\}$
$=\{\cos \frac{ C }{2}\}\{ a \frac{\sin \frac{ B }{2}}{\sin \frac{A}{2}}+ b \frac{\sin \frac{ A }{2}}{\sin \frac{B}{2}}\}$
$=\sqrt{\frac{s(s-c)}{a b}}\{a \frac{\sqrt{\frac{(s-a)(s-c)}{a c}}}{\sqrt{\frac{(s-b)(s-c)}{b c}}}+b \frac{\sqrt{\frac{(s-b)(s-c)}{b c}}}{\sqrt{\frac{(s-a)(s-c)}{a c}}}\}$
$=\sqrt{\frac{s(s-c)}{a b}}\{\sqrt{(\frac{s-a}{s-b}) a b}+\sqrt{(\frac{s-b}{s-a}) a b}\}$
$=\sqrt{s(s-c)}\{\frac{s-a+s-b}{\sqrt{(s-a)(s-b)}}\}$
$=\sqrt{s(s-c)}\{\frac{2 s-a-b}{\sqrt{(s-a)(s-b)}}\}$
$=c \sqrt{\frac{s(s-c)}{(s-a)(s-b)}}=c \cot \frac{C}{2}$
Alternate Method:
Let $a =1, b=\sqrt{3}, c =2$ and $A =30^{\circ}$,
$B =60^{\circ}, C =90^{\circ}$.
Hence, the given expression is equal to 2 , which is given by option (D).

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