Question

If $7{\sin }^2\theta +3{\cos }^2\theta =4,(0^o<\theta <{90}^o)$, then the value of ${\tan }^2\theta$ is:

• A. $\frac{1}{3}$
• B. $\frac{1}{2}$
• C. $1$
• D. $\sqrt{3}$

Answer 1

The correct option is C $1$

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

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

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

$\theta ={45}^0\Rightarrow (\tan 45{)}^2=1^2=1$

$\therefore {\tan }^2\theta =1$