Question

In $\Delta ABC$, If $a,b,c$ are in $H.P$., then ${\sin }^2\left(\frac{A}{2}\right),$ ${\sin }^2\left(\frac{B}{2}\right),$ ${\sin }^2\left(\frac{C}{2}\right)$ are in

• A. A.P.
• B. G.P.
• C. H.P.
• D. A.G.P.

Answer 1

Answer: (C)

H.P.

As we know $a,b,c$ are in $H.P.$

$\frac{1}{a}$ $,$ $\frac{1}{b}$ $,$ $\frac{1}{c}$ are in $A.P.$

Multiplying by $s$ in all terms, then also it will remain $A.P.$

$\therefore$$\frac{s}{a}$ $,$ $\frac{s}{b}$ $,$ $\frac{s}{c}$ are in $A.P.$

Subtracting by $1$ in all terms, then also it will remain $A.P.$

$\therefore$$\frac{s-a}{a}$ $,$ $\frac{s-b}{b}$ $,$ $\frac{s-b}{c}$ are in $A.P.$

$\therefore$$\frac{a}{s-a}$ $,$ $\frac{b}{s-b}$ $,$ $\frac{c}{s-c}$ are in $H.P.$

As we know by u\sing the half angle formula in sides form, The value of $\sin \left(\frac{A}{2}\right)$ $=$$\sqrt{\frac{(s-b)(s-c)}{bc}}$

$\therefore$ ${\sin }^2\left(\frac{A}{2}\right)$ $=$$\frac{(s-b)(s-c)}{bc}$

Similarly,

${\sin }^2\left(\frac{B}{2}\right)$ $=$$\frac{(s-a)(s-c)}{ac}$

${\sin }^2\left(\frac{C}{2}\right)$ $=$$\frac{(s-a)(s-b)}{ab}$

Multipying all the three terms by $abc$ and dividing each term by $(s-a)(s-b)(s-c)$, we get

$\frac{a}{s-a}$ $,$ $\frac{b}{s-b}$ $,$ $\frac{c}{s-c}$

And we know from above that$\frac{a}{s-a}$ $,$ $\frac{b}{s-b}$ $,$ $\frac{c}{s-c}$ are in $H.P.$

so, $\therefore$ ${\sin }^2\left(\frac{A}{2}\right),$ ${\sin }^2\left(\frac{B}{2}\right),$ ${\sin }^2\left(\frac{C}{2}\right)$ are in $H.P.$