Question

In $△ABC,(cot\frac{A}{2}+cot\frac{B}{2})(asi{n}^2\frac{B}{2}+bsi{n}^2\frac{A}{2})=$
[Roorkee 1988]

• A. cot C
• B. c cot C
• C. cot $\frac{C}{2}$
• D. c cot $\frac{C}{2}$

Answer 1

The correct option is D c cot $\frac{C}{2}$

$\{cot\frac{A}{2}+cot\frac{B}{2}\}\{asi{n}^2\frac{B}{2}+bsi{n}^2\frac{A}{2}\}$
$=\left\{\frac{cot\frac{C}{2}}{sin\frac{A}{2}sin\frac{B}{2}}\right\}\{asi{n}^2\frac{B}{2}+bsi{n}^2\frac{A}{2}\}$
$=\left\{cos\frac{C}{2}\right\}\{a\frac{sin\frac{B}{2}}{sin\frac{A}{2}}+b\frac{sin\frac{A}{2}}{sin\frac{B}{2}}\}$
$=\sqrt{\frac{s(s-c)}{ab}}\{a\frac{\sqrt{\frac{(s-a)(s-c)}{ac}}}{\sqrt{\frac{(s-b)(s-c)}{bc}}}+b\frac{\sqrt{\frac{(s-b)(s-c)}{bc}}}{\sqrt{\frac{(s-a)(s-c)}{ac}}}\}$
$=\sqrt{\frac{s(s-c)}{ab}}\left\{\sqrt{\left(\frac{s-a}{s-b}\right)ab+}\sqrt{\left(\frac{s-b}{s-a}\right)ab}\right\}$
$=\sqrt{s(s-c)}\sqrt{\frac{s-a+s-b}{\sqrt{(s-a)(s-b)}}}=\left\{\sqrt{s(s-c)}\sqrt{\frac{2s-a-b}{\sqrt{(s-a)(s-b)}}}\right\}$
$=c\sqrt{\frac{s(s-c)}{(s-a)(s-b)}}=ccot\frac{C}{2}.$
Trick : Such type of unconditional problems can be checked by putting the particular values for a = 1, $b=\sqrt{3},$ c = 2 and A = 30^{o}, B = 60^{o}, C = 90^{o}. \)

Hence expression is equal to 2 which is given by (d).