Question

If $k=\tan {25}^o$ then find $\frac{k-1}{k+1}+\frac{k+1}{k-1}$.

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

Answer 1

The correct option is C $2\sec {130}^o$

$\frac{k-1}{k+1}+\frac{k+1}{k-1}$

$=\frac{(k-1{)}^2+(k+1{)}^2}{k^2-1}$

$=\frac{2(k^2+1)}{k^2-1}$

$=\frac{2\left({\tan }^2\right({25}^o)+1)}{{\tan }^2\left({25}^o\right)-1}$

$=\frac{2{\sec }^2\left({25}^o\right)}{\left(\frac{{\sin }^2\left({25}^o\right)-{\cos }^2\left({25}^o\right)}{{\cos }^2\left({25}^o\right)}\right)}$

$=\frac{2{\sec }^2\left({25}^o\right){\cos }^2\left({25}^o\right)}{{\sin }^2\left({25}^o\right)-{\cos }^2\left({25}^o\right)}$

$=\frac{2}{-\cos \left({50}^o\right)}$

$=\frac{2}{\cos \left({130}^o\right)}$

$=2\sec \left({130}^o\right)$