Question

If $A+B+C=180°$, then the value of $cot\frac{A}{2}+cot\frac{B}{2}+cot\frac{C}{2}$ will be

• A. $2cot\frac{A}{2}cot\frac{B}{2}cot\frac{C}{2}$
• B. $4cot\frac{A}{2}cot\frac{B}{2}cot\frac{C}{2}$
• C. $cot\frac{A}{2}cot\frac{B}{2}cot\frac{C}{2}$
• D. $8cot\frac{A}{2}cot\frac{B}{2}cot\frac{C}{2}$

Answer 1

The correct option is C

$cot\frac{A}{2}cot\frac{B}{2}cot\frac{C}{2}$

Explanation for the correct option:

Step 1: Finding cot of the given angle

Given, $A+B+C=180°$

Divide by 2

$\begin{array}{rcl}\frac{A}{2}+\frac{B}{2}+\frac{C}{2}& =& \frac{180°}{2}\\ \frac{A}{2}+\frac{B}{2}& =& \frac{180°}{2}-\frac{C}{2}\end{array}$

Taking “$cot$ ” both side,

$cot(\frac{A}{2}+\frac{B}{2})=cot(90°-\frac{C}{2})$
Using, $\mathit{c}\mathit{o}\mathit{t}\mathbf{(}\mathit{A}\mathbf{+}\mathit{B}\mathbf{)}\mathbf{=}\mathbf{}\frac{\mathbf{c}\mathbf{o}\mathbf{t}\mathbf{A}\mathbf{c}\mathbf{o}\mathbf{t}\mathbf{B}\mathbf{-}\mathbf{1}}{\mathbf{c}\mathbf{o}\mathbf{t}\mathbf{B}\mathbf{+}\mathbf{c}\mathbf{o}\mathbf{t}\mathbf{A}}$

Step2. Finding value for $\mathit{c}\mathit{o}\mathit{t}\frac{\mathbf{A}}{\mathbf{2}}\mathbf{+}\mathit{c}\mathit{o}\mathit{t}\frac{\mathbf{B}}{\mathbf{2}}\mathbf{+}\mathit{c}\mathit{o}\mathit{t}\frac{\mathbf{C}}{\mathbf{2}}$

$\begin{array}{ccc}\frac{cot{\displaystyle \frac{A}{2}}cot{\displaystyle \frac{B}{2}}–1}{cot{\displaystyle \frac{A}{2}}+cot{\displaystyle \frac{B}{2}}}& =& \tan \frac{C}{2}\\ \Rightarrow \frac{cot{\displaystyle \frac{A}{2}}cot{\displaystyle \frac{B}{2}}–1}{cot{\displaystyle \frac{A}{2}}+cot{\displaystyle \frac{B}{2}}}& =& \frac{1}{cot{\displaystyle \frac{c}{2}}}\\ \Rightarrow (cot\frac{A}{2}cot\frac{B}{2}–1)cot\frac{c}{2}& =& cot\frac{A}{2}+cot\frac{B}{2}\\ \Rightarrow cot\frac{A}{2}cot\frac{B}{2}cot\frac{c}{2}-cot\frac{c}{2}& =& cot\frac{A}{2}+cot\frac{B}{2}\\ \Rightarrow cot\frac{A}{2}cot\frac{B}{2}cot\frac{c}{2}& =& cot\frac{A}{2}+cot\frac{B}{2}+cot\frac{c}{2}\end{array}$

Hence, correct option is option(C) .