Question

If $\tan {20}^{\circ }=k\Rightarrow \frac{\tan {250}^{\circ }+\tan {340}^{\circ }}{\tan {200}^{\circ }-\tan {110}^{\circ }}=$

• A. $\frac{1-k^2}{1+k^2}$
• B. $\frac{1+k^2}{1-k^2}$
• C. $\frac{2k}{1-k^2}$
• D. $\frac{k+1}{k-1}$

Answer 1

The correct option is A $\frac{1-k^2}{1+k^2}$

$\tan {250}^{\circ }=\tan ({270}^{\circ }-{20}^{\circ })=\cot {20}^{\circ }$
$\tan {340}^{\circ }=\tan ({360}^{\circ }-{20}^{\circ })=-\tan {20}^{\circ }$
$\tan {200}^{\circ }=\tan ({180}^{\circ }+{20}^{\circ })=\tan {20}^{\circ }$
$\tan {110}^{\circ }=\tan ({90}^{\circ }+{20}^{\circ })=-\cot {20}^{\circ }$
Using these values in the given expression we get,
$\frac{\tan {250}^{\circ }+\tan {340}^{\circ }}{\tan {200}^{\circ }-\tan {110}^{\circ }}=\frac{\cot {20}^{\circ }-\tan {20}^{\circ }}{\tan {20}^{\circ }+\cot {20}^{\circ }}=\frac{\frac{1}{k}-k}{k+\frac{1}{k}}=\frac{1-k^2}{1+k^2}$