Question

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

• A. $\frac{2}{\sqrt{3}}-2$
• B. $\frac{2\sqrt{3}+6}{3}$
• C. $\frac{2}{\sqrt{3}}$
• D. $\sqrt{3}$

Answer 1

The correct option is B

$\frac{2\sqrt{3}+6}{3}$

$7{\sin }^2\theta +3{\cos }^2\theta =4\Rightarrow 7{\sin }^2\theta +3(1-{\sin }^2\theta )=4\Rightarrow 7si{n}^2\theta +3-3si{n}^2\theta =4\Rightarrow 4si{n}^2\theta =1\Rightarrow si{n}^2\theta =\frac{1}{4}\Rightarrow \sin \theta =\frac{1}{2}\therefore \theta ={30}^{\circ }\therefore sec{30}^{\circ }+cosec{30}^{\circ }=\left(\frac{2}{\sqrt{3}}\right)+2$

= $\frac{2+2\sqrt{3}}{\sqrt{3}}$
=$\frac{2(1+\sqrt{3})}{\sqrt{3}}$
Multiplying numerator and denominator by $\sqrt{3}$ , we get
= $\frac{2\sqrt{3}+6}{3}$