Question

In any $∆ABC$, $\frac{\tan {\displaystyle \frac{A}{2}}-\tan {\displaystyle \frac{B}{2}}}{\tan {\displaystyle \frac{A}{2}}+\tan {\displaystyle \frac{B}{2}}}$ is equal to:

• A. $\frac{a-b}{a+b}$
• B. $\frac{a-b}{c}$
• C. $\frac{c}{a-b}$
• D. None of these.

Answer 1

The correct option is B

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

Explanation for the correct option.

Step 1: Simplify $\frac{\tan {\displaystyle \frac{A}{2}}-\tan {\displaystyle \frac{B}{2}}}{\tan {\displaystyle \frac{A}{2}}+\tan {\displaystyle \frac{B}{2}}}$.

$$\begin{gathered} \begin{array}{rcl}\frac{\tan {\displaystyle \frac{A}{2}}-\tan {\displaystyle \frac{B}{2}}}{\tan {\displaystyle \frac{A}{2}}+\tan {\displaystyle \frac{B}{2}}}& =& \frac{{\displaystyle \frac{\sin A/2}{\cos A/2}}-{\displaystyle \frac{\sin B/2}{\cos B/2}}}{{\displaystyle \frac{\sin A/2}{\cos A/2}}+{\displaystyle \frac{\sin B/2}{\cos B/2}}}\\ & =& \frac{\sin {\displaystyle \frac{A}{2}}·\cos {\displaystyle \frac{B}{2}}-\sin {\displaystyle \frac{B}{2}}·\cos {\displaystyle \frac{B}{2}}}{\sin \frac{A}{2}·\cos \frac{B}{2}+\sin \frac{B}{2}·\cos \frac{B}{2}}\\ & =& \frac{\sin \left({\displaystyle \frac{A-B}{2}}\right)}{\sin \left({\displaystyle \frac{A+B}{2}}\right)}\left[\because sin(x-y)=sinxcosy-sinycosxand \\ sin\left(x+y\right)=sinxcosy+sinycosxand\right]\end{array} \end{gathered}$$

By angle sum property of a triangle,

$$\begin{gathered} A+B+C=180° \\ \Rightarrow \frac{A+B}{2}=90°-\frac{C}{2} \end{gathered}$$

So,

$\frac{\sin \left({\displaystyle \frac{A-B}{2}}\right)}{\sin \left({\displaystyle 90°-\frac{C}{2}}\right)}=\frac{\sin \left({\displaystyle \frac{A-B}{2}}\right)}{\cos \left({\displaystyle \frac{C}{2}}\right)}\mathbf{[}\mathbf{By}\mathbf{}\mathbf{sin}\mathbf{}\mathbf{}\left(90°-\theta \right)\mathbf{}\mathbf{=}\mathbf{}\mathbf{cos}\mathbf{}\mathbf{\theta }\mathbf{]}...\left(1\right)$

Multiplying the numerator and denominator of $\left(1\right)$ by $2\sin \frac{C}{2}$, we get

$\begin{array}{rcl}\frac{2\sin \left({\displaystyle \frac{C}{2}}\right)\sin \left({\displaystyle \frac{A-B}{2}}\right)}{2\sin \left({\displaystyle \frac{C}{2}}\right)\cos {\displaystyle \frac{C}{2}}}& =& \frac{2\sin \left({\displaystyle 90°-\frac{A+B}{2}}\right)\sin \left({\displaystyle \frac{A-B}{2}}\right)}{\sin 2\left({\displaystyle \frac{C}{2}}\right)}\left[\because sin2\theta =2sin\theta cos\theta \right]\\ & =& \frac{2\cos {\displaystyle \frac{A+B}{2}}\sin {\displaystyle \frac{A-B}{2}}}{\sin C}\left[\because sin\left(90°-\theta \right)=cos\theta \right]\\ & =& \frac{\sin A-\sin B}{\sin C}...\left(2\right)\left[\because sinx-siny=2cos\frac{x+y}{2}sin\frac{x-y}{2}\right]\end{array}$

We know that $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$, so $\left(2\right)$ will be

$$\begin{gathered} =\frac{{\displaystyle \frac{a}{k}}-{\displaystyle \frac{b}{k}}}{{\displaystyle \frac{c}{k}}} \\ =\frac{a-b}{c} \end{gathered}$$

Hence, option B is correct.