Question

If $0<\theta <{90}^o$ and $\sec \theta =\text{cosec}{60}^o$ , find the value of $2{\cos }^2\theta -1$

Answer 1

$0<\theta <{90}^{\circ }$

$sec\theta =cosec60$

$\Rightarrow sec\theta =\frac{1}{sin60}=\frac{2}{\sqrt{3}}$

$\Rightarrow \frac{1}{cos\theta }=\frac{2}{\sqrt{3}}\Rightarrow cos\theta =\frac{\sqrt{3}}{2}\Rightarrow \theta ={30}^{\circ }$

$2co{s}^2\theta -1=2(cos30{)}^2-1$

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

$\Rightarrow 2\times \frac{3}{4}-1\Rightarrow \frac{3}{2}-1=\frac{1}{2}$

$\therefore 2co{s}^2\theta -1=\frac{1}{2}$