Question

If $7{\sin }^2\theta +3{\cos }^2\theta =4$, show that $\tan \theta =\frac{1}{\sqrt{3}}$

Answer 1

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

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

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

$A{C}^2=A{B}^2+B{C}^2$

$4-1=AB\Rightarrow AB=\sqrt{3}$

$\tan \theta =\frac{BC}{BA}=\frac{1}{\sqrt{3}}$