Question

If the sides of the triangle $\Delta$$ABC$ $a,b,c$ are in $A.P$. then the value of $\tan \frac{A}{2}+\tan \frac{C}{2}$ is?

• A. $\frac{2}{3}\tan \frac{B}{2}$
• B. $\frac{2}{3}\cot \frac{B}{2}$
• C. $\frac{1}{4}\tan \frac{B}{2}$
• D. $\frac{1}{3}\cot \frac{B}{2}$

Answer 1

The correct option is B $\frac{2}{3}\cot \frac{B}{2}$

We know that , a,b,c are in AP

so,$\frac{a+c}{2}=b\Rightarrow a+c=2b\Rightarrow s=\frac{a+b+c}{2}=\frac{b+2b}{2}=\frac{3b}{2}.............................\left(1\right)$

From the Above formulaes,

$\tan \frac{A}{2}+tan\frac{C}{2}=\frac{△}{s(s-a)}+\frac{△}{s(s-c)}=\frac{△(2s-c-a)}{s(s-a)(s-c))}$

as, $2s-c-a=a+b+c-a-c=b$

$=\frac{△b(s-b)}{s(s-a)(s-b)(s-c)}=\frac{△(s-b)b}{{△}^2}=\frac{(s-b)b}{△}$

Now, Again from above Formulaes

$\tan \frac{B}{2}=\frac{△}{s(s-b)}\Rightarrow \cot \frac{B}{2}=\frac{s(s-b)}{△}$

So, now

$(\tan \frac{A}{2}+\tan \frac{C}{2}):\cot \frac{B}{2}=\frac{\frac{(s-b)b}{△}}{\frac{s(s-b)}{△}}=\frac{b}{s}=\frac{b}{\frac{3}{2}}=\frac{2}{3}\Rightarrow \tan \frac{A}{2}+\tan \frac{C}{2}=\frac{2}{3}\cot \frac{B}{2}$

Therefore, Answer is $B$