Question

Find the value of $\left(\frac{\tan {65}^o}{\cot {25}^o}\right)$

Answer 1

$\frac{tan{65}^{\circ }}{cot{25}^{\circ }}$

$=\frac{tan{65}^{\circ }}{tan(90-25{)}^{\circ }}:-(cot=tan/90-0)$

$\therefore =\frac{tan{65}^{\circ }}{tan{65}^{\circ }}=1$.