Question

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

Answer 1

Given that,

$A+B+C=\pi$ and $\cos A=\cos B\cos C$

$B+C=\pi -A$ …… (1)

On taking $\cos$ both side and we get

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

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

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

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

For Ist part,

$\frac{\mathrm{cosBcosC}-sinBsinC}{cosBcosC}=-1$

$\frac{\cos B\cos C}{\cos B\cos C}-\frac{sinBsinC}{cosBcosC}=-1$

$1-\tan B\tan C=-1$

$1+1=\mathrm{tanB\; tanC}$

$\tan B\tan C=2$

Taking reciprocal both side and we get,

$\frac{1}{tanBtanC}=\frac{1}{2}$

$\mathrm{cotBcotC}=\frac{1}{2}$

Hence, proved.

For IInd part,

L.H.S.

$\tan B+\tan C$

$=\frac{\sin B}{\cos B}+\frac{\sin C}{\cos C}$

$=\frac{\sin B\cos C+\cos B\sin C}{\mathrm{cosB}\cos C}$

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

$=\frac{\sin (\pi -A)}{\mathrm{cosA}}$

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

$=\tan AR.H.S.$

Hence, proved