Question

If $\theta$ is a positive acute angle such that $\sec \theta =\csc {60}^{\circ }$, find the value of $2{\cos }^2\theta -1.$

Answer 1

Given,

$\theta$ is acute and $\sec \theta =\csc {60}^{\circ }$

$\csc {60}^{\circ }=\frac{2}{\sqrt{3}}$

$\therefore \sec \theta =\frac{2}{\sqrt{3}}$

We know that, $\sec {30}^{\circ }=\frac{2}{\sqrt{3}}$

$⟹\sec \theta =\sec {30}^{\circ }$

$⟹\theta ={30}^{\circ }$

To Evaluate,

$2{\cos }^2\theta -1$

$=2{\cos }^2{30}^{\circ }-1$

$=2(\frac{\sqrt{3}}{2}{)}^2-1$

$=2\left(\frac{3}{4}\right)-1$

$=\frac{3}{2}-1$

$=\frac{1}{2}$