Question

If $A+B+C=\pi$ and $\cos A=\cos B.\cos C$ then show that $\cot B.\cot C=\frac{1}{2}$ and $\tan B+\tan C=\tan A$

Answer 1

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

$\therefore \cos A=\cos B.\cos C$

$\Rightarrow \cos (\pi -(B+C))=\cos B.\cos C$

$\Rightarrow -\cos (B+C)=\cos B.\cos C(\because \cos (\pi -A)=-\cos A)$

$\Rightarrow \cos (B+C)+\cos B.\cos C=0$

$\Rightarrow \cos B\cos C-\sin B\sin C+\cos B\cos C=0(\because \cos (A+B)=\cos A\cos B-\sin A\sin B)$

$\Rightarrow 2\cos B\cos C=\sin B\sin C$

$\Rightarrow \frac{2\cos B\cos C}{\sin B\sin C}=1(\because \cot \theta =\frac{\cos \theta }{\sin \theta })$

$\Rightarrow \cot B\cot C=\frac{1}{2}$

$\therefore \tan B\tan C=2(\because \cot \theta =\frac{1}{\tan \theta })$

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

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

$\tan A=\frac{-(\tan B+\tan C)}{1-2}$

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