Question

If $\tan \theta =\frac{3}{4}$. find the value of $\frac{1-\cos \theta }{1+\cos \theta }$

Answer 1

Given,

$\tan \theta =\frac{3}{4}$

$\Rightarrow ta{n}^2\theta =\frac{9}{16}$

$\Rightarrow se{c}^2\theta =1+\frac{9}{16}=\frac{25}{16}$
$\Rightarrow co{s}^2\theta =\frac{16}{25}$

$\Rightarrow \cos \theta =\frac{4}{5}$

$\frac{1-\cos \theta }{1+\cos \theta }$

$=\frac{1-\frac{4}{5}}{1+\frac{4}{5}}$..............(replacing value of $cos\theta$ )

$=\frac{1}{9}$