Question

If $\cot \theta =\frac{1}{\sqrt{3}}$ then find the value of $\left(\frac{1-{\cos }^2\theta }{2-{\sin }^2\theta }\right)$

Answer 1

Given,

$\cot \theta =\frac{1}{\sqrt{3}}\Rightarrow \cot \theta =\cot {60}^o\Rightarrow \theta ={60}^o\therefore \left(\frac{1-{\cos }^2\theta }{2-{\sin }^2\theta }\right)$

Putting the value of $\theta$ we get,

$=\frac{1-{\cos }^2{60}^o}{2-{\sin }^2{60}^o}=\frac{1-{\left(\frac{1}{2}\right)}^2}{2-{\left(\frac{\sqrt{3}}{2}\right)}^2}=\frac{1-\frac{1}{4}}{2-\frac{3}{4}}=\frac{\frac{4-1}{4}}{\frac{8-3}{4}}=\frac{\frac{3}{4}}{\frac{5}{4}}=\frac{3}{5}.$