Question

If $a,b,c$ are H.P. then ${\sin }^2\frac{A}{2},{\sin }^2\frac{B}{2},{\sin }^2\frac{C}{2}$ are in

• A. A.P.
• B. G.P.
• C. H.P.
• D. none of these

Answer 1

Answer: (C)

H.P.

Since $a,b,c$ are in $H.P.,$

$\therefore \frac{1}{b}-\frac{1}{a}=\frac{1}{c}-\frac{1}{b}$

$\frac{a-b}{ab}=\frac{b-c}{bc}$

using Sine rule

$\Rightarrow \frac{(\sin A-\sin B)}{\sin A\sin B}=\frac{\sin B-\sin C}{\sin B\sin C}.$

$\Rightarrow \frac{2\sin \frac{A-B}{2}\cos \frac{B+C}{2}}{2\sin \frac{A}{2}\cos \frac{A}{2}}=\frac{2\sin \frac{B-C}{2}\cos \frac{B+C}{2}}{2\sin \frac{C}{2}\cos \frac{C}{2}}$

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

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

$\Rightarrow {\sin }^2\frac{C}{2}({\sin }^2\frac{A}{2}-{\sin }^2\frac{B}{2})={\sin }^2\frac{A}{2}({\sin }^2\frac{B}{2}-{\sin }^2\frac{C}{2})$

Divide by ${\sin }^2\frac{A}{2}{\sin }^2\frac{B}{2}{\sin }^2\frac{C}{2}$

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

$\therefore \frac{1}{{\sin }^2\frac{A}{2}},\frac{1}{{\sin }^2\frac{B}{2}},\frac{1}{{\sin }^2\frac{C}{2}}$ are in $A.P.$

$\Rightarrow {\sin }^2\frac{A}{2},{\sin }^2\frac{B}{2},{\sin }^2\frac{C}{2}$ are in $H.P.$