Question

If $\tan \theta =\frac{1}{\sqrt{7}}$ for $\pi <\theta <\frac{3\pi }{2}$ then find the value of $\frac{{\csc }^2\theta -{\sec }^2\theta }{{\csc }^2\theta +{\sec }^2\theta }$

Answer 1

$\begin{array}{c}\tan \theta =\frac{1}{\sqrt{7}}=\frac{1}{6}\\ \frac{\cos e{c}^2\theta -{\sec }^2\theta }{\cos e{c}^2\theta +{\sec }^2\theta }\\ \frac{{\left(h/p\right)}^2-{\left(h/b\right)}^2}{{\left(h/p\right)}^2+{\left(h/b\right)}^2}\\ \frac{8-8/7}{8+8/7}\\ \frac{\left(56-8\right)/7}{\left(56+8\right)/7}=\\ \frac{48}{7}\times \frac{7}{64}\\ \frac{48}{64}\\ \frac{16\times 3}{16\times 4}\\ \frac{3}{4}\end{array}$