Question

$\frac{\tan {57}^{\circ }+\cot {37}^{\circ }}{\tan {33}^{\circ }+\cot {53}^{\circ }}$ is equal to

• A. $\tan {33}^{\circ }\cot {57}^{\circ }$
• B. $\tan {57}^{\circ }\cot {37}^{\circ }$
• C. $\tan {33}^{\circ }\cot {53}^{\circ }$
• D. $\tan {53}^{\circ }\cot {37}^{\circ }$

Answer 1

The correct option is B $\tan {57}^{\circ }\cot {37}^{\circ }$

Given: $\frac{\tan {57}^o+\cot {37}^o}{\tan {33}^o+\cot {53}^o}$

$\Rightarrow \frac{\tan {57}^o+\cot {37}^o}{\tan (90-57)+\cot (90-37{)}^o}$

$\Rightarrow \frac{\tan {57}^o+\cot {37}^o}{\cot {57}^o+\tan {37}^o}$

$\Rightarrow \frac{\tan {57}^o+\cot {37}^o}{\frac{1}{\tan 57}+\frac{1}{\cot {37}^o}}$

$\Rightarrow \tan {57}^o.\cot {37}^o\left[\frac{\tan {57}^o+\cot {37}^o}{\tan {57}^o+\cot {37}^o}\right]$

$\Rightarrow \tan {57}^o\cot {37}^o$