Question

If $\cot \theta =\frac{1}{\sqrt{3}}$ then prove that
$\frac{1-{\cos }^2\theta }{2-{\sin }^2\theta }=\frac{3}{5}$.

Answer 1

Given, $\cot \theta =\frac{1}{\sqrt{3}}$
$\cot \theta =\cot {60}^o$
$\therefore \theta ={60}^o$
Now, $\frac{1-{\cos }^2\theta }{2-{\sin }^2\theta }=\frac{1-{\cos }^2{60}^o}{2-{\sin }^2{60}^o}$
$\frac{1-{\left(\frac{1}{2}\right)}^2}{2-{\left(\frac{\sqrt{3}}{2}\right)}^2}$
$=\frac{1-\frac{1}{4}}{2-\frac{3}{4}}=\frac{3}{4}\times \frac{4}{5}=\frac{3}{5}$
Thus, $\frac{1-{\cos }^2\theta }{2-{\sin }^2\theta }=\frac{3}{5}$