Question

If θ is an acute angel such that $se{c}^2\theta =3,$ then find the value of
$\frac{{\tan }^2\theta -cose{c}^2\theta }{{\tan }^2\theta +cose{c}^2\theta }$

• A. 1/14
• B. 1/7
• C. 1
• D. 0

Answer 1

The correct option is B 1/7

We have,

$\theta$ is an acute angle such that ${\sec }^2\theta =3$

$\Rightarrow {\tan }^2\theta ={\sec }^2\theta -1=3-1=2$

Now,

${\tan }^2\theta =\frac{1}{{\cot }^2\theta }\Rightarrow {\cot }^2\theta =\frac{1}{2}$

Also, $cose{c}^2\theta ={\cot }^2\theta +1=1+\frac{1}{2}=\frac{3}{2}$

Hence,

$\frac{{\tan }^2\theta -cose{c}^2\theta }{{\tan }^2\theta +cose{c}^2\theta }=\frac{2-\frac{3}{2}}{2+\frac{3}{2}}=\frac{1}{7}$