Question

If $7si{n}^2\theta +3co{s}^2\theta =4$, then $sec\theta +cosec\theta$ is equal to

• A. $\frac{2}{\sqrt{3}}-2$
• B. $\frac{2}{\sqrt{3}}+2$
• C. $\frac{2}{\sqrt{3}}$
• D. None of these

Answer 1

The correct option is B

$\frac{2}{\sqrt{3}}+2$

$7{\sin }^2\theta +3{\cos }^2\theta =4\Rightarrow 7{\sin }^2\theta +3(1-{\sin }^2\theta )=4[\because co{s}^2\theta +si{n}^2\theta =1]\Rightarrow 7si{n}^2\theta +3-3si{n}^2\theta =4\Rightarrow 4si{n}^2\theta =1\Rightarrow si{n}^2\theta =\frac{1}{4}\Rightarrow \sin \theta =\frac{1}{2}\therefore \theta ={30}^{\circ }$

We have,
$sec{30}^{\circ }=\frac{2}{\sqrt{3}}andcosec{30}^{\circ }=2\therefore sec{30}^{\circ }+cosec{30}^{\circ }=\frac{2}{\sqrt{3}}+2$