Question

In a $\Delta ABC,\frac{{\cos }^2\left(\frac{B-C}{2}\right)}{(b+c{)}^2}+\frac{{\sin }^2\left(\frac{B-C}{2}\right)}{(b-c{)}^2}=$

• A. $1/a^2$
• B. $2/a^2$
• C. $3/a^2$
• D. $4/a^2$

Answer 1

Answer: (A)

$\frac{{\cos }^2\left(\frac{B-C}{2}\right)}{(b+c{)}^2}+\frac{{\sin }^2\left(\frac{B-C}{2}\right)}{(b-c{)}^2}$
$\left\{\begin{array}{c}\text{ by using sine theorem }\\ \frac{\sin B}{b}=\frac{\sin C}{c}=\frac{1}{k}\text{ (let) }\end{array}\right\}$
$(b+c{)}^2=k^2(\sin B+\sin C{)}^2$
$=4k^2{\sin }^2\left(\frac{B+C}{2}\right){\cos }^2\left(\frac{B-C}{2}\right)$
$\Rightarrow 4k^2{\sin }^2\left(\frac{180-A}{2}\right){\cos }^2\left(\frac{B-C}{2}\right)$
$\Rightarrow \frac{(b+c{)}^2}{{\cos }^2\left(\frac{B-C}{2}\right)}=4k^2{\cos }^2\left(\frac{A}{2}\right)$
$\Rightarrow \frac{1}{4k^2{\cos }^2\left(\frac{A}{2}\right)}+\frac{1}{4k^2{\sin }^2\left(\frac{A}{2}\right)}$
$\Rightarrow \frac{{\cos }^2\left(\frac{A}{2}\right)+{\sin }^2\left(\frac{A}{2}\right)}{4k^2{\cos }^2\frac{A}{2}{\sin }^2\frac{A}{2}}$
$\Rightarrow \frac{1}{{\left(2k{\sin }^2\cos \frac{A}{2}\right)}^2}\Rightarrow \frac{1}{k^2{\sin }^{2}A}=\frac{1}{a^2}$
$a=k{\sin }^A$
$SO,\frac{1}{a^2}$