Question

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

Answer 1

(i) $A=180-(B+C)$

$\cos A=\cos (180-(B_C\left)\right)$

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

Given $\cos A=\cos B\cos C$

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

$\cos B\cos C=-\cos B\cos C+\sin B\sin C$

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

$\Rightarrow \frac{\cos B\cos C}{\sin B\sin C}=\frac{1}{2}$

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

Given $\cos A=\cos B\cos C$

(ii) $A+B+C=180$

$B+C=180-A$

$=\sin (B+C)=\sin (180-A)$

$=\sin (B+C)=\sin A$

Also, $\cos B.\cos C=\cos A$

$\therefore$ Consider,$\tan A-\tan B-\tan C$

$=\tan A-(\frac{\sin B}{\cos B}+\frac{\sin C}{\cos C})$

$=\tan A-\left[\frac{\sin B\cos C+\sin C\cos B}{\cos B\cos C}\right]$

$=\tan A-\frac{\left[\sin \right(B+C\left)\right]}{\cos C\cos C}$

$=\tan A-\frac{\sin (180-A)}{\cos A}$

$=\tan A-\frac{\sin A}{\cos A}=\tan A-\tan A$

$=0$

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