Question

`\text{Evaluate }\frac{\tan 65^\circ }{\cot 25^\circ}`

 

Answer 1

65° + 25° = 90°

`\therefore \frac{\tan 65^\circ }{\cot 25^\circ }=\frac{\tan(90^\circ -25^\circ )}{\cot25^\circ }=\frac{\cot 25^\circ}{\cot 25^\circ }=1`