Question

Evaluate $\frac{\tan {65}^o}{\cot {25}^o}$

Answer 1

Consider $\tan {65}^0$ which can be rewritten as follows:

$\tan {65}^0=\tan {({90}^0-25}^0)=\cot {25}^0(\because \tan ({90}^0-x)=\cot x)$

Therefore, $\frac{\tan {65}^0}{\cot {25}^0}$ can be evaluated as shown below:

$\frac{\tan {65}^0}{\cot {25}^0}=\frac{\cot {25}^0}{\cot {25}^0}=1$

Hence, $\frac{\tan {65}^0}{\cot {25}^0}=1$.