Question

In a $\Delta ABC,$ $\frac{1}{1+{\tan }^2\frac{A}{2}}+\frac{1}{1+{\tan }^2\frac{B}{2}}+\frac{1}{1+{\tan }^2\frac{C}{2}}=k[1+\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}],$ then the value of $k$ is

Answer 1

$\text{LHS}=\frac{1}{1+{\tan }^2\frac{A}{2}}+\frac{1}{1+{\tan }^2\frac{B}{2}}+\frac{1}{1+{\tan }^2\frac{C}{2}}$
$={\cos }^2\frac{A}{2}+{\cos }^2\frac{B}{2}+{\cos }^2\frac{C}{2}$
$=\frac{1+\cos A}{2}+\frac{1+\cos B}{2}+\frac{1+\cos C}{2}$
$=\frac{3}{2}+\frac{1}{2}(\cos A+\cos B+\cos C)$
$=\frac{3}{2}+\frac{1}{2}[1-2{\sin }^2\frac{A}{2}+2\cos \left(\frac{B+C}{2}\right)\cos \left(\frac{B-C}{2}\right)]$
$=\frac{3}{2}+\frac{1}{2}[1-2{\sin }^2\frac{A}{2}-2\sin \frac{A}{2}\cos \left(\frac{B-C}{2}\right)]$
$=2-\sin \frac{A}{2}[\sin \frac{A}{2}-\cos \left(\frac{B-C}{2}\right)]$
$=2-\sin \frac{A}{2}[\cos \left(\frac{B+C}{2}\right)-\cos \left(\frac{B-C}{2}\right)]$
$=2+2\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}$
$=2[1+\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}]$
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$\text{RHS}=k[1+\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}]$
$\therefore k=2$


Alternate solution:
If we assume $A=B=C={60}^{\circ }$, we get
$\frac{1}{1+{\tan }^2\frac{A}{2}}+\frac{1}{1+{\tan }^2\frac{B}{2}}+\frac{1}{1+{\tan }^2\frac{C}{2}}$
$={\cos }^2\frac{A}{2}+{\cos }^2\frac{B}{2}+{\cos }^2\frac{C}{2}={\cos }^2{30}^{\circ }+{\cos }^2{30}^{\circ }+{\cos }^2{30}^{\circ }=3\times \frac{3}{4}=\frac{9}{4}$

$k[1+\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}]=k[1+(\sin {30}^{\circ }{)}^3]=k[1+\frac{1}{8}]=\frac{9k}{8}$
Now,
$\frac{9k}{8}=\frac{9}{4}\therefore k=2$