Question

What is $\frac{\cot {54}^o}{\tan {36}^o}+\frac{\tan {20}^o}{\cot {70}^o}$ equal to?

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is C $2$

We know that, $\tan \theta =\cot ({90}^o-\theta )$

So, $\frac{\cot {54}^o}{\tan {36}^o}+\frac{\tan {20}^o}{\cot {70}^o}=\frac{\tan (90-54)}{\tan {36}^o}+\frac{\tan {20}^o}{\tan (90-70)}=\frac{\tan {36}^o}{\tan {36}^o}+\frac{\tan {20}^o}{\tan {20}^o}=1+1=2$