Question

If ${sin}^2\theta +5{cos}^2\theta =4,$ then find $\theta$ and hence prove that $sec\theta +cosec\theta =2+\frac{2}{\sqrt{3}}$

Answer 1

Given, ${sin}^2\theta +5{cos}^2\theta =4,$

$\Rightarrow 1-{\cos }^2\theta +5{\cos }^2\theta =4$

$\Rightarrow 4{\cos }^2\theta$ = $3$

Solving we get $\cos \theta =\frac{\sqrt{3}}{2}$

So $\theta ={30}^{\circ }$

Now $\sec \theta +cosec\theta$ = $\sec (30{)}^{\circ }+cosec(30{)}^{\circ }=$ $\frac{2}{\sqrt{3}}+2$