Solveeit Logo
Login

Question

Question: If \(A + B + C = 180^{o},\) then the value of \(\cot\frac{A}{2} + \cot\frac{B}{2} + \cot\frac{C}{2}\...

If A+B+C=180o,A + B + C = 180^{o}, then the value of cot⁡A2+cot⁡B2+cot⁡C2\cot\frac{A}{2} + \cot\frac{B}{2} + \cot\frac{C}{2} will be

A

2cot⁡A2cot⁡B2cot⁡C22\cot\frac{A}{2}\cot\frac{B}{2}\cot\frac{C}{2}

B

4cot⁡A2cot⁡B2cot⁡C24\cot\frac{A}{2}\cot\frac{B}{2}\cot\frac{C}{2}

C

cot⁡A2cot⁡B2cot⁡C2\cot\frac{A}{2}\cot\frac{B}{2}\cot\frac{C}{2}

D

8cot⁡A2cot⁡B2cot⁡C28\cot\frac{A}{2}\cot\frac{B}{2}\cot\frac{C}{2}

Answer

cot⁡A2cot⁡B2cot⁡C2\cot\frac{A}{2}\cot\frac{B}{2}\cot\frac{C}{2}

Explanation

Solution

A+B+C=180oA + B + C = 180^{o} ∴A2+B2=90o−C/2\therefore\frac{A}{2} + \frac{B}{2} = 90^{o} - C/2

∴\therefore cot⁡(A2+B2)=cot⁡(90o−C2)\cot\left( \frac{A}{2} + \frac{B}{2} \right) = \cot\left( 90^{o} - \frac{C}{2} \right)

or cot⁡A2.cot⁡B2−1cot⁡B2+cot⁡A2=tan⁡C2=1cot⁡C2\frac{\cot\frac{A}{2}.\cot\frac{B}{2} - 1}{\cot\frac{B}{2} + \cot\frac{A}{2}} = \tan\frac{C}{2} = \frac{1}{\cot\frac{C}{2}}

or(cot⁡A2.cot⁡B2−1)cot⁡C2=cot⁡B2+cot⁡A2\left( \cot\frac{A}{2}.\cot\frac{B}{2} - 1 \right)\cot\frac{C}{2} = \cot\frac{B}{2} + \cot\frac{A}{2}; cot⁡A2.cot⁡B2.cot⁡C2=cot⁡C2+cot⁡B2+cot⁡A2\cot\frac{A}{2}.\cot\frac{B}{2}.\cot\frac{C}{2} = \cot\frac{C}{2} + \cot\frac{B}{2} + \cot\frac{A}{2}