Question

Find the magnitude of angle $A$, if:

$\tan A-2\cos A\tan A+2\cos A-1=0$

Answer 1

$\tan A-2\cos A\tan A+2\cos A-1=0$
$\frac{\sin A}{\cos A}-2\cos A\cdot \frac{\sin A}{\cos A}+2\cos A-1=0$
$\frac{\sin A-2\sin A\cos A+2{\cos }^{2}A-\cos A}{\cos A}=0$
$\sin A-2\cos A(\sin A-\cos A)-\cos A=0$$[\cos A\ne 0]$
$(\sin A-\cos A)-2\cos A(\sin A-\cos A)=0$
$(\sin A-\cos A)(1-2\cos A)=0$
$\therefore \sin A-\cos A=0$ or $(1-2\cos A)=0$
$\sin A=\cos A⟹\frac{\sin A}{\cos A}=1$
$\tan A=1⟹A={45}^0$
$1-2\cos A=0$
$2\cos A=1⟹\cos A=\frac{1}{2}$
$\therefore A={60}^0$