Question

Given that $\tan \theta =\frac{1}{\sqrt{3}}$, find the value of $\frac{\cos e{c}^2\theta -{\sec }^2\theta }{\cos e{c}^2\theta +{\sec }^2\theta }$

Answer 1

Given that

$\tan \theta =1/\sqrt{3}$

$\Rightarrow \tan \theta =\tan {30}^o$

$\Rightarrow \theta ={30}^o$

Now,

$\frac{cose{c}^2\theta -{\sec }^2\theta }{cose{c}^2\theta +{\sec }^2\theta }$

$=\frac{cose{c}^2{30}^o-{\sec }^2{30}^o}{cose{c}^2{30}^o+{\sec }^2{30}^o}$

$=\frac{(2{)}^2-(2/\sqrt{3}{)}^2}{(2{)}^2+(2/\sqrt{3}{)}^2}$

$=\frac{4-4/3}{4+4/3}$

$=\frac{12-4/3}{12+4/3}$

$=\frac{8}{16}$

$=\frac{1}{2}$

$\therefore \frac{cose{c}^2\theta -{\sec }^2\theta }{cose{c}^2\theta +{\sec }^2\theta }=\frac{1}{2}$.