Question

In a $△ABC$, Prove that $\cot A\cot B+\cot B\cot C+\cot C\cot A=1$

Answer 1

L.H.S

$=\cot A\cot B+\cot B\cot C+\cot C\cot A$

$=\frac{1}{\tan A\tan B}+\frac{1}{\mathrm{tanB}\mathrm{tanC}}+\frac{1}{\tan C\tan A}$

$=\frac{1}{\tan A\tan B}+\frac{1}{\mathrm{tanB}\mathrm{tanC}}+\frac{1}{\tan C\tan A}$

$=\frac{tanA+tanB+\mathrm{tanC}}{\tan A\tan B\tan C}$ …….. (1)

We know that

$A+B+C=\pi$

$A+B=\pi -C$

On taking tan both sides, we get

$\tan (A+B)=\tan (\pi -C)$

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

$\tan A+\tan B=-\tan C+\tan A\tan B\tan C$

$\tan A+\tan B+\tan C=\tan A\tan B\tan C$

From equation (1), we get

$=1$

R.H.S

Hence. proved.