Question

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

Answer 1

To prove

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

R.H.S

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

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

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

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

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

L.H.S

Hence, proved.