Question

If in triangle $A B C$, $a^{2}+2 b c-\left(b^{2}+c^{2}\right)=a b \sin \frac{C}{2} \cos \frac{C}{2}$, then $\cot (B+C)=$

• A. $-\frac{8}{15}$
• B. $\frac{1}{4}$
• C. $-\frac{15}{8}$
• D. 4

Answer 1

Answer: (C)

Given, $a^{2}+2 b c-\left(b^{2}+c^{2}\right)=a b \sin \frac{C}{2} \cos \frac{C}{2}$
$\Rightarrow \, a^{2}+2 b c-\left(b^{2}+c^{2}\right)=\frac{1}{2} a b \sin C$
$\Rightarrow \, a^{2}-\left(b^{2}+c^{2}-2 b c\right)=\frac{1}{2} a b \sin C$
$\Rightarrow \, a^{2}-(b-c)^{2}=\frac{1}{2} a b \sin C$
$\Rightarrow \,(a -b +c)(a +b -c)=\frac{1}{2} a b \sin C$
$4(s-b)(s-c)=\Delta$
$\Rightarrow \, 4=\frac{\sqrt{s(s-a)(s-b)(s-c)}}{(s-b)(s-c)}$
$\Rightarrow \, 4=\sqrt{\frac{s(s-a)}{(s-b)(s-c)}}$
$\Rightarrow \cot \frac{A}{2}=4$
$\Rightarrow \tan \frac{A}{2}=\frac{1}{4}$
$\therefore \tan A=\frac{2 \tan \frac{A}{2}}{1-\tan ^{2} \frac{A}{2}}$
$\Rightarrow \, \tan A=\frac{2\left(\frac{1}{4}\right)}{1-\frac{1}{16}}=\frac{8}{15}$
$\Rightarrow \cot A=\frac{15}{8}$
$\Rightarrow \cot [\pi-(B+C)]=\frac{15}{8}$
$\Rightarrow \cot (B+C)=-\frac{15}{8}$