Question

If $7{\sin }^2\theta +3{\cos }^2\theta =4$ then prove that $\sec \theta +\text{cosec}\theta =2+\frac{2}{\sqrt{3}}$.

Answer 1

$7{\sin }^2\theta +3{\cos }^2\theta =4$

$\Rightarrow 7{\sin }^2\theta +3(1-{\sin }^2\theta )=4$

$\Rightarrow 7{\sin }^2\theta +3-3{\sin }^2\theta =4$

$\Rightarrow 4{\sin }^2\theta =1$

$\Rightarrow {\sin }^2\theta =\frac{1}{4}$

$\Rightarrow \sin \theta =\frac{1}{2}$

$\Rightarrow \sin \theta =\sin {30}^{\circ }$

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

L.H.S $=\sec \theta +\text{cosec}\theta$

$=\sec {30}^{\circ }+\text{cosec}{30}^{\circ }$

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

$=$ R.H.S

Hence proved.