Question

If $k=\tan {25}^o$, then $\frac{k-1}{k+1}+\frac{k+1}{k-1}$ is equal to:

• A. $2cosec{130}^0$
• B. $2cosec{45}^o$
• C. $2\sec {130}^o$
• D. $2\sec {400}^o$

Answer 1

Answer: (C)

$2\sec {130}^o$

The given expression is
$\frac{\tan {25}^{\circ }-1}{\tan {25}^{\circ }+1}+\frac{\tan {25}^{\circ }+1}{\tan {25}^{\circ }-1}$

$=\frac{\tan {25}^{\circ }-\tan {45}^{\circ }}{\tan {25}^{\circ }\tan {45}^{\circ }+1}+\frac{\tan {25}^{\circ }+\tan {45}^{\circ }}{\tan {25}^{\circ }\tan {45}^{\circ }-1}$

$=\tan (-{20}^{\circ })-\tan \left({70}^{\circ }\right)$

$=-(\tan {20}^{\circ }+\tan {70}^{\circ })$

$=-(\frac{\sin {20}^{\circ }}{\cos {20}^{\circ }}+\frac{\sin {70}^{\circ }}{\cos {70}^{\circ }})$

$=-\frac{1}{\cos {20}^{\circ }\sin {20}^{\circ }}$

$=-\frac{2}{\sin {40}^o}=\frac{2}{\cos (90+40{)}^o}=2\sec {130}^o$