Question

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

Answer 1

$\frac{1+\cos \theta }{1-\cos \theta }=\frac{16}{9}$

$9+9\cos \theta =16-16\cos \theta$

$25\cos \theta =7$

$\cos \theta =\frac{7}{25}$

$\angle ACB=\theta$

$\cos \theta =\frac{BC}{AC}=\frac{7}{25}$

${AB}^2+{BC}^2={AC}^2$

${AB}^2={25}^2-7^2$

$AB=24$

$\cot \theta =\frac{\cos \theta }{\sin \theta }=\frac{BC}{AC}\times \frac{AC}{AB}$

$=\frac{BC}{AB}=\frac{7}{24}$

$\frac{1+\cos \theta }{1-\cos \theta }=\frac{1+\frac{7}{24}}{1-\frac{7}{24}}=\frac{24+7}{24-7}=\frac{31}{17}$