In $\triangle \mathrm{ABC}$ if $\mathrm{a}, \mathrm{b}, \mathrm{c}_{,}$are in A.P. then $\cot \frac{A}{2}, \cot \frac{B}{2}, \cot \frac{C}{2}$ are also in A.P.
Now, $\cot \frac{A}{2}+\cot \frac{C}{2}=\frac{\cos \frac{A}{2}}{\sin \frac{A}{2}}+\frac{\cos \frac{C}{2}}{\sin \frac{C}{2}}$
$=\frac{\cos \frac{A}{2} \cdot \sin \frac{C}{2}+\sin \frac{A}{2} \cdot \cos \frac{C}{2}}{\sin \frac{A}{2} \cdot \sin \frac{C}{2}}$
$=\frac{\sin \left(\frac{\mathrm{A}}{2}+\frac{\mathrm{C}}{2}\right)}{\sin \frac{\mathrm{A}}{2} \cdot \sin \frac{\mathrm{C}}{2}}$
$=\frac{\sin \left(\frac{\pi}{2}-\frac{B}{2}\right)}{\sqrt{\frac{(s-b)(s-c)}{b c}} \cdot \sqrt{\frac{(s-a)(s-b)}{a b}}}$
$\cdots[\because \mathrm{A}+\mathrm{B}+\mathrm{C}=\pi]$
$=\frac{\cos \frac{B}{2}}{\frac{(s-b)}{b} \cdot \sqrt{\frac{(s-c)(s-a)}{c a}}}$
$=\frac{b \cos \frac{\mathrm{B}}{2}}{(s-b) \cdot \sin \frac{\mathrm{B}}{2}}$
$=\frac{b}{s-b} \cdot \cot \frac{B}{2}$
$=\frac{b}{\left(\frac{a+b+c}{2}-b\right)} \cdot \cot \frac{\mathrm{B}}{2}$
$\cdots[\because 2 s=a+b+c]$
$=\frac{2 b}{(2 b-b)} \cdot \cot \frac{\mathrm{B}}{2} \quad \ldots[$ By $(1)]$
$=\frac{2 b}{b} \cdot \cot \frac{B}{2}$
$=\left(\frac{2 b}{a+c-b}\right) \cdot \cot \frac{\mathrm{B}}{2}$
$\therefore \cot \frac{A}{2}+\cot \frac{C}{2}=2 \cot \frac{B}{2}$
Hence, $\cot \frac{A}{2}, \cot \frac{B}{2}, \cot \frac{C}{2}$ are in A.P.
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