Question

Find the value of $\frac{\sin {18}^o}{\cos {72}^o}\times \frac{\tan {26}^o}{\cot {64}^o}$.

Answer 1

Now,

$\frac{\sin {18}^o}{\cos {72}^o}\times \frac{\tan {26}^o}{\cot {64}^o}$

$=\frac{\sin ({90}^o-{72}^o)}{\cos {72}^o}\times \frac{\tan ({90}^o-{64}^o)}{\cot {64}^o}$

$=\frac{\cos \left({72}^o\right)}{\cos {72}^o}\times \frac{\cot \left({64}^o\right)}{\cot {64}^o}$

$=1\times 1$

$=1$.