Question

If $\tan {20}^o=p$, then $\frac{\tan {160}^o-\tan {110}^o}{1+\tan {160}^o\tan {110}^o}$ is equal to:

• A. $\left(\frac{1-p^2}{2p}\right)$
• B. $\left(\frac{2p}{1+p^2}\right)$
• C. $\left(\frac{1+p}{2p}\right)$
• D. $\left(\frac{1-p}{2p}\right)$

Answer 1

Answer: (A)

$\left(\frac{1-p^2}{2p}\right)$

$\frac{\tan {160}^o-\tan {110}^o}{1+\tan {160}^o\tan {110}^o}$

$=\tan ({160}^o-{110}^o)=\tan {50}^o=\cot {40}^o=\frac{1}{\tan {40}^o}$

$=\frac{1}{\frac{2\tan {20}^o}{1-{\tan }^2{20}^o}}=\frac{1-p^2}{2p}$