Question

Given that $16\cot A=12$; find the value of $\frac{\sin A+\cos A}{\sin A-\cos A}$.

Answer 1

$\cot A=\frac{12}{16}=\frac{3}{4}\Rightarrow \tan A=\frac{4}{3}\frac{\sin A+\cos A}{\sin A-\cos A}=\frac{\cos A(1+\tan A)}{\cos A(-1+\tan A)}=-\frac{(1+\tan A)}{(1-\tan A)}=\frac{-(1+\frac{4}{3})}{(1-\frac{4}{3})}=-\left(\frac{7}{-1}\right)=7$