Question

If $\cos A=m\cos B$, then

• A. $cot\frac{(A+B)}{2}=\frac{(m+1)}{(m–1)}\times \tan \frac{(B–A)}{2}$
• B. $\tan \frac{(A+B)}{2}=\frac{(m+1)}{(m–1)}\times cot\frac{(B–A)}{2}$
• C. $cot\frac{(A+B)}{2}=\frac{(m+1)}{(m–1)}\times \tan \frac{(A-B)}{2}$
• D. None of these

Answer 1

The correct option is A

$cot\frac{(A+B)}{2}=\frac{(m+1)}{(m–1)}\times \tan \frac{(B–A)}{2}$

The explanation for the correct option:

Step 1. Given that $\begin{array}{rcl}\cos A& =& m\cos B\end{array}$

$\Rightarrow$$\begin{array}{rcl}\frac{\cos A}{\cos B}& =& \frac{m}{1}\end{array}$

Using componendo and dividendo rule,

$\frac{(\cos A+\cos B)}{(\cos A–\cos B)}=\frac{(m+1)}{(m–1)}$

Step 3. Using the formulas

$$\begin{gathered} \mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{x}\mathbf{}\mathbf{+}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{y}\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\left(\frac{x+y}{2}\right)\mathbf{}\mathit{c}\mathit{o}\mathit{s}\left(\frac{x–y}{2}\right)and \\ \mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{x}\mathbf{}\mathbf{–}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{y}\mathbf{}\mathbf{=}\mathbf{}\mathbf{-}\mathbf{}\mathbf{2}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\left(\frac{x+y}{2}\right)\mathbf{}\mathit{s}\mathit{i}\mathit{n}\left(\frac{x–y}{2}\right)\mathbf{,} \end{gathered}$$

$\left[\frac{2\cos \left({\displaystyle \frac{A+B}{2}}\right)\cos \left({\displaystyle \frac{A–B}{2}}\right)}{-2\sin \left(\frac{A+B}{2}\right)\sin \frac{(A–B)}{2}}\right]=\frac{(m+1)}{(m–1)}$

$\Rightarrow$$\left[\frac{\cos \left({\displaystyle \frac{A+B}{2}}\right)\cos \left({\displaystyle \frac{B-A}{2}}\right)}{\sin \left(\frac{A+B}{2}\right)\sin \frac{(B-A)}{2}}\right]=\frac{(m+1)}{(m–1)}$ $\left[\mathbf{\because }\mathit{s}\mathit{i}\mathit{n}\mathit{c}\mathit{e}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{(}\mathbf{-}\mathit{x}\mathbf{)}\mathbf{=}\mathit{c}\mathit{o}\mathit{s}\mathit{x}\right]$

$\Rightarrow$ $cot\left(\frac{A+B}{2}\right)cot\left(\frac{B–A}{2}\right)=\frac{(m+1)}{(m–1)}$

$\Rightarrow$ $cot\left(\frac{A+B}{2}\right)=\left[\frac{(m+1)}{(m–1)}\right]/cot\left(\frac{B–A}{2}\right)$

$\mathbf{\therefore }\mathit{c}\mathit{o}\mathit{t}\mathbf{\left(}\frac{\mathbf{A}\mathbf{+}\mathbf{B}}{\mathbf{2}}\mathbf{\right)}\mathbf{=}\frac{\mathbf{(}\mathbf{m}\mathbf{+}\mathbf{1}\mathbf{)}}{\mathbf{(}\mathbf{m}\mathbf{–}\mathbf{1}\mathbf{)}}\mathbf{}\mathbf{\times }\mathbf{}\mathit{t}\mathit{a}\mathit{n}\mathbf{\left(}\frac{\mathbf{B}\mathbf{–}\mathbf{A}}{\mathbf{2}}\mathbf{\right)}$

Hence, the correct answer is option A