If the sides a, b and c of ∆ABC are in H.P.,


If the sides $a, b$ and $c$ of $\triangle A B C$ are in H.P., prove that $\sin ^{2} \frac{A}{2}, \sin ^{2} \frac{B}{2}$ and $\sin ^{2} \frac{C}{2}$ are in H.P.


$\sin ^{2} \frac{A}{2}, \sin ^{2} \frac{B}{2}$ and $\sin ^{2} \frac{C}{2}$ is a H.P.

$\Leftrightarrow \frac{1}{\sin ^{2} \frac{A}{2}}, \frac{1}{\sin ^{2} \frac{B}{2}}$ and $\frac{1}{\sin ^{2} \frac{C}{2}}$ is an A.P.

$\Leftrightarrow \frac{1}{\sin ^{2} \frac{B}{2}}-\frac{1}{\sin ^{2} \frac{A}{2}}=\frac{1}{\sin ^{2} \frac{C}{2}}-\frac{1}{\sin ^{2} \frac{B}{2}}$

$\Leftrightarrow \frac{\sin ^{2} \frac{A}{2}-\sin ^{2} \frac{B}{2}}{\sin ^{2} \frac{A}{2} \sin ^{2} \frac{B}{2}}=\frac{\sin ^{2} \frac{B}{2}-\sin ^{2} \frac{C}{2}}{\sin ^{2} \frac{B}{2} \sin ^{2} \frac{C}{2}}$

$\Leftrightarrow \frac{\sin \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)}{\sin ^{2} \frac{A}{2}}=\frac{\sin \left(\frac{B+C}{2}\right) \sin \left(\frac{B-C}{2}\right)}{\sin ^{2} \frac{C}{2}}$

$\Leftrightarrow \frac{\cos \left(\frac{C}{2}\right) \sin \left(\frac{A-B}{2}\right)}{\sin ^{2} \frac{A}{2}}=\frac{\cos \left(\frac{A}{2}\right) \sin \left(\frac{B-C}{2}\right)}{\sin ^{2} \frac{C}{2}} \quad[\mathrm{As}, A+B+C=\pi]$

$\Leftrightarrow \sin ^{2} \frac{C}{2} \cos \left(\frac{C}{2}\right) \sin \left(\frac{A-B}{2}\right)=\sin ^{2} \frac{A}{2} \cos \left(\frac{A}{2}\right) \sin \left(\frac{B-C}{2}\right)$

$\Leftrightarrow 2 \sin \frac{C}{2} \sin \frac{C}{2} \cos \left(\frac{C}{2}\right) \sin \left(\frac{A-B}{2}\right)=2 \sin \frac{A}{2} \sin \frac{A}{2} \cos \left(\frac{A}{2}\right) \sin \left(\frac{B-C}{2}\right)$

$\Leftrightarrow \sin \frac{C}{2} \sin C \sin \left(\frac{A-B}{2}\right)=\sin \frac{A}{2} \sin A \sin \left(\frac{B-C}{2}\right) \quad[\because \sin 2 \theta=2 \sin \theta \cos \theta]$

$\Leftrightarrow \sin C \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)=\sin A \cos \left(\frac{B+C}{2}\right) \sin \left(\frac{B-C}{2}\right) \quad[$ As,$A+B+C=\pi]$

$\Leftrightarrow \sin C \frac{(\sin A-\sin B)}{2}=\sin A \frac{(\sin B-\sin C)}{2} \quad\left[\sin C-\sin D=2 \cos \left(\frac{C+D}{2}\right) \sin \left(\frac{C-D}{2}\right)\right]$

$\Leftrightarrow \sin C(\sin A-\sin B)=\sin A(\sin B-\sin C)$

$\Leftrightarrow c k(a k-b k)=a k(b k-c k) \quad\left(\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=k\right.$ (say) $)$

$\Leftrightarrow c a-c b=a b-a c$

$\Leftrightarrow 2 a c=a b+b c$

$\Leftrightarrow \frac{2}{b}=\frac{1}{c}+\frac{1}{a} \quad$ [multiplying both the sides by $a b c$ ]

$\Leftrightarrow a, b, c$ are in H.P.

Leave a comment