Question

If in a $∆\mathrm{ABC}$, $a\tan A+b\tan B=(a+b)\tan \frac{1}{2}(A+B)$, then

• A. $A=B=C$
• B. $C=A$
• C. $A=B$
• D. $B=C$

Answer 1

The correct option is C

$A=B$

Explanation for the correct option.

Find the relation:

Given,

$a\tan A+b\tan B=(a+b)\tan \frac{1}{2}(A+B)$.

$\begin{array}{rcl}& & \\ & \Rightarrow & a\tan A-a\tan \frac{1}{2}(A+B)=b\tan \frac{1}{2}(A+B)-b\tan B\\ & \Rightarrow & a\left[\tan A-\tan \frac{1}{2}(A+B)\right]=b\left[\tan \frac{1}{2}(A+B)-\tan B\right]\\ & \Rightarrow & a\left[\frac{\sin A}{\cos A}-\frac{\sin \frac{1}{2}(A+B)}{\cos \frac{1}{2}(A+B)}\right]=b\left[\frac{\sin \frac{1}{2}(A+B)}{\cos \frac{1}{2}(A+B)}-\frac{\sin B}{\cos B}\right]\\ & \Rightarrow & a\left[\frac{\sin A\times \cos \frac{1}{2}(A+B)-\sin \frac{1}{2}(A+B)\times \cos A}{\cos A\times \cos \frac{1}{2}(A+B)}\right]=b\left[\frac{\sin \frac{1}{2}(A+B)\times \cos B-\sin B\times \cos \frac{1}{2}(A+B)}{\cos \frac{1}{2}(A+B)\times \cos B}\right]\\ & \Rightarrow & a\left[\sin \left(A-\frac{1}{2}(A+B)\right)\right]=b\left[\sin \left(\frac{1}{2}(A+B)-B\right)\right]\\ & \Rightarrow & a\left[\sin \left(\frac{A-B}{2}\right)\right]=b\left[\sin \left(\frac{A-B}{2}\right)\right]\\ & \Rightarrow & a=b\end{array}$

Since, the sine rule is,

$\frac{a}{\sin A}=\frac{b}{sinB}=\frac{c}{\sin C}$

Then,

Since, $a=b$, then

$$\begin{gathered} \sin A=\sin B \\ \Rightarrow A=B \end{gathered}$$

Hence, the correct option is C.