Question

The value of $\frac{\tan \frac{1}{2}A}{(a-b)(a-c)}+\frac{\tan \frac{1}{2}B}{(b-c)(b-a)}+\frac{\tan \frac{1}{2}C}{(c-a)(c-b)}$ is equal to

• A. $\frac{1}{△}$
• B. $\frac{2}{△}$
• C. $△$
• D. $noneofthese$

Answer 1

The correct option is A $\frac{1}{△}$

Let

$P=\frac{\tan \left(A/2\right)}{(a-b)(a-c)}+\frac{\tan \left(B/2\right)}{(b-c)(b-a)}+\frac{\tan \left(C/2\right)}{(c-a)(c-b)}$

Consider $△ABC$

with $a=2$ (hypotenuse)

$\angle A=90°$

$b=1$ & $c=\sqrt{3}$

So, $△=\frac{1}{2}b\times c=\frac{1}{2}\times \sqrt{3}\times 1=\frac{\sqrt{3}}{2}$

Now, $\tan B=\frac{b}{c}=\frac{1}{\sqrt{3}}$

So, $B=30°$

So, $C=60°$

$\tan 15°=\tan \left(\frac{B}{2}\right)=\tan (45°-30°)=-\left[\frac{1-\sqrt{3}}{1+\sqrt{3}}\right]=\frac{\sqrt{3}-1}{\sqrt{3}+1}$

So, $P=\frac{\tan 45°}{(2-1)(2-\sqrt{3})}+\frac{\tan 15°}{(1-\sqrt{3})(1-2)}+\frac{\tan 30°}{(\sqrt{3}-2)(\sqrt{3}-1)}$

$P=\frac{1}{2-\sqrt{3}}+\frac{1}{\sqrt{3}+1}+\frac{1}{\sqrt{3}}\times \frac{1}{(\sqrt{3}-2)(\sqrt{3}-1)}$

$P=\frac{\sqrt{3}+1+2-\sqrt{3}}{(2-\sqrt{3})(\sqrt{3}+1)}+\frac{1}{\sqrt{3}}\times \frac{1}{(\sqrt{3}-2)(\sqrt{3}-1)}$

$P=\frac{3}{(2-\sqrt{3})(\sqrt{3}+1)}-\frac{1}{\sqrt{3}}\frac{1}{(2-\sqrt{3})(\sqrt{3}-1)}$

$P=\frac{3\sqrt{3}(\sqrt{3}-1)-(\sqrt{3}+1)}{\sqrt{3}(2-\sqrt{3})(\sqrt{3}+1)(\sqrt{3}-1)}$

$P=\frac{9-3\sqrt{3}-\sqrt{3}-1}{\sqrt{3}\left(2\right)(2-\sqrt{3})}=\frac{9(2-\sqrt{3})}{2\sqrt{3}(2-\sqrt{3})}$

$P=\frac{1}{\sqrt{3}/2}=\frac{1}{△}$