Question

If $\cot \theta =\frac{1}{\sqrt{3}}$, show that $\frac{1-co{s}^2\theta }{2-{\sin }^2\theta }=\frac{3}{5}$

Answer 1

Given,

$cot\theta =\frac{1}{\sqrt{3}}$

$\tan \theta =\frac{1}{\cot \theta }=\frac{1}{\frac{1}{\sqrt{3}}}=\sqrt{3}$

From Pythagoras theorem,

$A{C}^2=A{B}^2+B{C}^2$

$A{C}^2=(\sqrt{3}{)}^2+1^2$

$B{C}^2=3+1=4$

$AC=2$

$\sin \theta =\frac{oppositeSide}{Hypotenuse}=\frac{\sqrt{3}}{2}$

$\cos \theta =\frac{AdjacentSide}{Hypotenuse}=\frac{1}{2}$

Therefore,

$\frac{1-{\cos }^2\theta }{2-{\sin }^2\theta }=\frac{1-{\left(\frac{1}{2}\right)}^2}{2-{\left(\frac{\sqrt{3}}{2}\right)}^2}$

$=\frac{(4-1)}{(8-3)}$

$=\frac{3}{5}$

$\therefore \frac{1-{\cos }^2\theta }{2-{\sin }^2\theta }=\frac{3}{5}$