Question

If$A+B+C=180$ then prove that $\cot B\cot C+\cot C\cot A+\cot A\cot B$=1

Answer 1

$\Rightarrow$$A+B=180-C$

Now take $tan$ of both sides

$\Rightarrow$$\frac{\left(\tan A+\tan B\right)}{\left(1-\tan A\tan B\right)}=-\tan C$

$\Rightarrow$$\tan A+\tan B+\mathrm{tanC}=tanAtanBtanC$

And, dividing both sides by $tanAtanBtanC$

we get

$\Rightarrow$$\frac{1}{\tan B}\times \frac{1}{\tan C}+\frac{1}{\tan A}\times \frac{1}{\tan C}+\frac{1}{\tan A}\times \frac{1}{\tan B}=1$

$\Rightarrow$$\cot B\times \cot C+\cot A\times \cot C+\cot A\times \cot B=1$

Hence proved.