Question

If $\tan \theta =\frac{12}{5}$. find the value of $\frac{1+\sin \theta }{1-\sin \theta }$

Answer 1

we have,

$\tan \theta =\frac{a}{b}$

$c=\sqrt{a^2+b^2}$

$\Rightarrow \sin \theta =\frac{a}{c}$

Given,

$\tan \theta =\frac{12}{5}$

$c^2={12}^2+5^2$

$\Rightarrow c=13$

$\sin \theta =\frac{12}{13}$

$\frac{1+\sin \theta }{1-\sin \theta }=\frac{1+\frac{12}{13}}{1-\frac{12}{13}}$

$=\frac{25}{1}=25$