A geometry problem by A Former Brilliant Member

Geometry Level 2

If $\cos \theta=\dfrac{\sqrt{3}}{2}$ , then find the value of $x$ if $x=1-\tan^2 \theta$ .

\dfrac{2}{3}

-\dfrac{1}{3}

\dfrac{3}{2}

\dfrac{1}{3}

1 solution

A Former Brilliant Member
Jul 12, 2017

If $cos \theta=\dfrac{\sqrt{3}}{2}$ , then $\sec \theta=\dfrac{2}{\sqrt{3}}$ .

$x=1-\tan^2 \theta$ , however, $\tan^2 \theta=\sec^2 \theta - 1$ , therefore

$x=1-(\sec^2 \theta - 1)=1-\sec^2 \theta +1=2-\sec^2 \theta$ , however, $\sec \theta=\dfrac{2}{\sqrt{3}}$ , therefore

$x=2-\sec^2 \theta=2-\left(\dfrac{2}{\sqrt{3}}\right)^2=2-\dfrac{4}{3}=\dfrac{6}{3}-\dfrac{4}{3}=$ $\boxed{\dfrac{2}{3}}$