Question

If $\tan \theta =\frac{1}{\sqrt{7}}$, show that $\frac{cose{c}^2\theta -{\sec }^2\theta }{cose{c}^2\theta +{\sec }^2\theta }=\frac{3}{4}$.

Answer 1

$\tan \theta =\frac{1}{\sqrt{7}},\frac{\sin \theta }{\cos \theta }=\frac{1}{\sqrt{7}},\frac{{\sin }^2\theta }{{\cos }^2\theta }=\frac{1}{7},{\cos }^2\theta +{\sin }^2\theta =1$

${\cos }^2=7{\sin }^2\theta ,7{\sin }^2\theta +{\sin }^2\theta =1\Rightarrow =\frac{1}{8}\therefore {\cos }^2\theta =1-{\sin }^2\theta =1-\frac{1}{8}=\frac{7}{8}\therefore cose{c}^2\theta =\frac{1}{{\sin }^2\theta }=8\therefore {\sec }^2=\frac{1}{{\cos }^2\theta }=\frac{8}{7}\therefore \frac{cose{c}^2\theta -{\sec }^2\theta }{cose{c}^2\theta +{\sec }^2\theta }=\frac{8-\frac{8}{7}}{8+\frac{8}{7}}=\frac{3}{4}$

Henceproved