Question

If $\tan 2A=\cot (A-{18}^o)$ then find the value of A where (2A) and $(A-{18}^o)$ are acute angles

Answer 1

$\tan 2A=\cot (A-18)$
$=>\cot (90-2A)=\cot (A-18)$
$=>90-2A=A-18$
$=>3A=108$
$=>A={36}^{\circ }$