Question

Find the value of : $\frac{2\tan {53}^o}{\cot {37}^o}-\frac{\cot {80}^o}{\tan {10}^o}$

Answer 1

$\frac{2\tan {53}^o}{\cot {37}^o}-\frac{\cot {80}^o}{\tan {10}^o}$

= $\frac{2\tan (90-37{)}^o}{\cot {37}^o}-\frac{\cot (90-10{)}^o}{\tan {10}^o}$

= $\frac{2\cot {37}^o}{\cot {37}^o}-\frac{\tan {10}^o}{\tan {10}^o}$

= $2-1$
= $1$