Question

If $\frac{1+\cos \theta }{1-\cos \theta }=\frac{16}{9}$, find $\frac{1+\cot \theta }{1-\cot \theta }$

• A. $\frac{31}{17}$
• B. $\frac{17}{31}$
• C. $\frac{31}{37}$
• D. $\frac{19}{17}$

Answer 1

The correct option is A $\frac{31}{17}$

$1+cos\theta =2co{s}^2\left(\frac{\theta }{2}\right)$
$1-cos\theta =2si{n}^2\left(\frac{\theta }{2}\right)$
Hence substituting in the above equation, we get
$co{t}^2\frac{\theta }{2}=\frac{16}{9}$
$cot\left(\frac{\theta }{2}\right)=\frac{4}{3}$
$tan\left(\frac{\theta }{2}\right)=\frac{3}{4}$
Now
$tan\left(\theta \right)=\frac{2tan\frac{\theta }{2}}{1-ta{n}^2\frac{\theta }{2}}$
$=\frac{\frac{3}{2}}{1-\frac{9}{16}}$
$=\frac{24}{7}$
Therefore
$\frac{1+cot\theta }{1-cot\theta }$
$=\frac{24+7}{24-7}$
$=\frac{31}{17}$